Figure P8.63. Figure P8.70. Figure P8.72. Figure P8.67. Figure P8.65. Figure P8.68. Figure P8.66. Figure P8.71. Figure P8.74

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1 248 CHAPTER 8 Potential Energy and Conservation of Energy Probles 249 R T v T v B B Figure P An object of ass is suspended fro a post on top of a cart by a string of length L as in Figure P8.65a. The cart and object are initially oving to the right at constant speed v i. The cart coes to rest after colliding and sticking to a buper as in Figure P8.65b, and the suspended object swings through an angle. (a) Show that the speed is v i 2gL( cos ). (b) If L.20 and 35.0, find the initial speed of the cart. (Hint: The force eerted by the string on the object does no work on the object.) v i L (a) Figure P A child slides without friction fro a height h along a curved water slide (Fig. P8.66). She is launched fro a height h/5 into the pool. Deterine her aiu airborne height y in ters of h and. θ (b) k 67. A ball having ass is connected by a strong string of length L to a pivot point and held in place in a vertical position. A wind eerting constant force of agnitude F is blowing fro left to right as in Figure P8.67a. (a) If the ball is released fro rest, show that the aiu height H it reaches, as easured fro its initial height, is H 2L (g/f ) 2 Check that the above forula is valid both when 0 H L and when L H 2L. (Hint: First deterine the potential energy associated with the constant wind force.) (b) Copute the value of H using the values 2.00 kg, L 2.00, and F 4.7 N. (c) Using these sae values, deterine the equilibriu height of the ball. (d) Could the equilibriu height ever be greater than L? Eplain. F (a) Pivot L F Figure P8.67 Pivot 68. A ball is tied to one end of a string. The other end of the string is fied. The ball is set in otion around a vertical circle without friction. At the top of the circle, the ball has a speed of v i Rg, as shown in Figure P8.68. At what angle should the string be cut so that the ball will travel through the center of the circle? The path after string is cut R θ C (b) v i = Rg L H than the tension at the top by a value si ties the weight of the ball. 70. A pendulu coprising a string of length L and a sphere swings in the vertical plane. The string hits a peg located a distance d below the point of suspension (Fig. P8.70). (a) Show that if the sphere is released fro a height below that of the peg, it will return to this height after striking the peg. (b) Show that if the pendulu is released fro the horizontal position ( 90 ) and is to swing in a coplete circle centered on the peg, then the iniu value of d ust be 3L/5. Figure P Jane, whose ass is 50.0 kg, needs to swing across a river (having width D) filled with an-eating crocodiles to save Tarzan fro danger. However, she ust swing into a wind eerting constant horizontal force F on a vine having length L and initially aking an angle with the vertical (Fig. P8.7). Taking D 50.0, F 0 N, L 40.0, and 50.0, (a) with what iniu speed ust Jane begin her swing to just ake it to Wind F Tarzan φ θ L θ Peg L d Jane the other side? (Hint: First deterine the potential energy associated with the wind force.) (b) nce the rescue is coplete, Tarzan and Jane ust swing back across the river. With what iniu speed ust they begin their swing? Assue that Tarzan has a ass of 80.0 kg. 72. A child starts fro rest and slides down the frictionless slide shown in Figure P8.72. In ters of R and H, at what height h will he lose contact with the section of radius R? H Figure P A 5.00-kg block free to ove on a horizontal, frictionless surface is attached to one end of a light horizontal spring. The other end of the spring is fied. The spring is copressed 0.00 fro equilibriu and is then released. The speed of the block is.20 /s when it passes the equilibriu position of the spring. The sae eperient is now repeated with the frictionless surface replaced by a surface for which k Deterine the speed of the block at the equilibriu position of the spring. 74. A 50.0-kg block and a 00-kg block are connected by a string as in Figure P8.74. The pulley is frictionless and of negligible ass. The coefficient of kinetic friction between the 50.0-kg block and the incline is k Deterine the change in the kinetic energy of the 50.0-kg block as it oves fro to, a distance of R h Figure P8.66 θ y h/5 Figure P A ball at the end of a string whirls around in a vertical circle. If the ball s total energy reains constant, show that the tension in the string at the botto is greater D Figure P kg 37.0 Figure P kg v

2 250 CHAPTER 8 Potential Energy and Conservation of Energy ANSWERS T QUICK QUIZZES 8. Yes, because we are free to choose any point whatsoever as our origin of coordinates, which is the U g 0 point. If the object is below the origin of coordinates that we choose, then U g 0 for the object Earth syste. 8.2 Yes, the total echanical energy of the syste is conserved because the only forces acting are conservative: the force of gravity and the spring force. There are two fors of potential energy: () gravitational potential energy and (2) elastic potential energy stored in the spring. 8.3 The first and third balls speed up after they are thrown, while the second ball initially slows down but then speeds up after reaching its peak. The paths of all three balls are parabolas, and the balls take different ties to reach the ground because they have different initial velocities. However, all three balls have the sae speed at the oent they hit the ground because all start with the sae kinetic energy and undergo the sae change in gravitational potential energy. In other words, E total 2 v 2 gh is the sae for all three balls at the start of the otion. 8.4 Designate one object as No. and the other as No. 2. The eternal force does work W app on the syste. If W app 0, then the syste energy increases. If W app 0, then the syste energy decreases. The effect of friction is to decrease the total syste energy. Equation 8.5 then becoes E W app E friction K U [K f K 2f ) (K i K 2i )] [(U gf U g 2f U sf ) (U gi U g 2i U si )] You ay find it easier to think of this equation with its ters in a different order, saying total initial energy net change total final energy K i K 2i U gi U g2i U si W app f k d K f K 2f U gf U g 2f U sf 8.5 The slope of a U()-versus- graph is by definition du()/d. Fro Equation 8.6, we see that this epression is equal to the negative of the coponent of the conservative force acting on an object that is part of the syste.

3 P U Z Z L E R Airbags have saved countless lives by reducing the forces eerted on vehicle occupants during collisions. How can airbags change the force needed to bring a person fro a high speed to a coplete stop? Why are they usually safer than seat belts alone? (Courtesy of Saab) 252 CHAPTER 9 Linear Moentu and Collisions Consider what happens when a golf ball is struck by a club. The ball is given a very large initial velocity as a result of the collision; consequently, it is able to travel ore than 00 through the air. The ball eperiences a large acceleration. Furtherore, because the ball eperiences this acceleration over a very short tie interval, the average force eerted on it during the collision is very great. According to Newton s third law, the ball eerts on the club a reaction force that is equal in agnitude to and opposite in direction to the force eerted by the club on the ball. This reaction force causes the club to accelerate. Because the club is uch ore assive than the ball, however, the acceleration of the club is uch less than the acceleration of the ball. ne of the ain objectives of this chapter is to enable you to understand and analyze such events. As a first step, we introduce the concept of oentu, which is useful for describing objects in otion and as an alternate and ore general eans of applying Newton s laws. For eaple, a very assive football player is often said to have a great deal of oentu as he runs down the field. A uch less assive player, such as a halfback, can have equal or greater oentu if his speed is greater than that of the ore assive player. This follows fro the fact that oentu is defined as the product of ass and velocity. The concept of oentu leads us to a second conservation law, that of conservation of oentu. This law is especially useful for treating probles that involve collisions between objects and for analyzing rocket propulsion. The concept of the center of ass of a syste of particles also is introduced, and we shall see that the otion of a syste of particles can be described by the otion of one representative particle located at the center of ass. Linear Moentu and Collisions c h a p t e r Definition of linear oentu of a particle 9. LINEAR MMENTUM AND ITS CNSERVATIN In the preceding two chapters we studied situations too cople to analyze easily with Newton s laws. In fact, Newton hiself used a for of his second law slightly different fro F a (Eq. 5.2) a for that is considerably easier to apply in coplicated circustances. Physicists use this for to study everything fro subatoic particles to rocket propulsion. In studying situations such as these, it is often useful to know both soething about the object and soething about its otion. We start by defining a new ter that incorporates this inforation: The linear oentu of a particle of ass oving with a velocity v is defined to be the product of the ass and velocity: p v (9.) Chapter utline 9. Linear Moentu and Its Conservation 9.2 Ipulse and Moentu 9.3 Collisions 9.4 Elastic and Inelastic Collisions in ne Diension 9.5 Two-Diensional Collisions 9.6 The Center of Mass 9.7 Motion of a Syste of Particles 9.8 (ptional) Rocket Propulsion Linear oentu is a vector quantity because it equals the product of a scalar quantity and a vector quantity v. Its direction is along v, it has diensions ML/T, and its SI unit is kg /s. If a particle is oving in an arbitrary direction, p ust have three coponents, and Equation 9. is equivalent to the coponent equations p v p y v y p z v z (9.2) As you can see fro its definition, the concept of oentu provides a quantitative distinction between heavy and light particles oving at the sae velocity. For eaple, the oentu of a bowling ball oving at 0 /s is uch greater than that of a tennis ball oving at the sae speed. Newton called the product v

4 9. Linear Moentu and Its Conservation CHAPTER 9 Linear Moentu and Collisions quantity of otion; this is perhaps a ore graphic description than our present-day word oentu, which coes fro the Latin word for oveent. Quick Quiz 9. Two objects have equal kinetic energies. How do the agnitudes of their oenta copare? (a) p p 2, (b) p p 2, (c) p p 2, (d) not enough inforation to tell. Because the tie derivative of the total oentu p tot p p 2 is zero, we conclude that the total oentu of the syste ust reain constant: p tot p p p 2 constant (9.4) syste or, equivalently, p i p 2i p f p 2f (9.5) 6.2 Using Newton s second law of otion, we can relate the linear oentu of a particle to the resultant force acting on the particle: The tie rate of change of the linear oentu of a particle is equal to the net force acting on the particle: (9.3) In addition to situations in which the velocity vector varies with tie, we can use Equation 9.3 to study phenoena in which the ass changes. The real value of Equation 9.3 as a tool for analysis, however, stes fro the fact that when the net force acting on a particle is zero, the tie derivative of the oentu of the particle is zero, and therefore its linear oentu is constant. f course, if the particle is isolated, then by necessity F 0 and p reains unchanged. This eans that p is conserved. Just as the law of conservation of energy is useful in solving cople otion probles, the law of conservation of oentu can greatly siplify the analysis of other types of coplicated otion. Conservation of Moentu for a Two-Particle Syste Consider two particles and 2 that can interact with each other but are isolated fro their surroundings (Fig. 9.). That is, the particles ay eert a force on each other, but no eternal forces are present. It is iportant to note the ipact of Newton s third law on this analysis. If an internal force fro particle (for eaple, a gravitational force) acts on particle 2, then there ust be a second internal force equal in agnitude but opposite in direction that particle 2 eerts on particle. Suppose that at soe instant, the oentu of particle is p and that of particle 2 is p 2. Applying Newton s second law to each particle, we can write where F 2 is the force eerted by particle 2 on particle and F 2 is the force eerted by particle on particle 2. Newton s third law tells us that F 2 and F 2 are equal in agnitude and opposite in direction. That is, they for an action reaction pair F 2 F 2. We can epress this condition as or as F 2 dp dt dp dt F dp dt dp 2 dt d(v) dt F 2 F 2 0 and F 2 dp 2 dt d dt (p p 2 ) 0 In this chapter, the ters oentu and linear oentu have the sae eaning. Later, in Chapter, we shall use the ter angular oentu when dealing with rotational otion. Newton s second law for a particle F 2 p = v 2 F 2 p 2 = 2 v 2 Figure 9. At soe instant, the oentu of particle is p v and the oentu of particle 2 is p 2 2 v 2. Note that F 2 F 2. The total oentu of the syste p tot is equal to the vector su p p 2. Conservation of oentu EXAMPLE 9. where p li and p 2i are the initial values and p f and p 2f the final values of the oentu during the tie interval dt over which the reaction pair interacts. Equation 9.5 in coponent for deonstrates that the total oenta in the, y, and z directions are all independently conserved: p i p f (9.6) syste syste p iy syste p fy syste p iz syste p fz syste This result, known as the law of conservation of linear oentu, can be etended to any nuber of particles in an isolated syste. It is considered one of the ost iportant laws of echanics. We can state it as follows: Whenever two or ore particles in an isolated syste interact, the total oentu of the syste reains constant. This law tells us that the total oentu of an isolated syste at all ties equals its initial oentu. Notice that we have ade no stateent concerning the nature of the forces acting on the particles of the syste. The only requireent is that the forces ust be internal to the syste. Quick Quiz 9.2 Your physical education teacher throws a baseball to you at a certain speed, and you catch it. The teacher is net going to throw you a edicine ball whose ass is ten ties the ass of the baseball. You are given the following choices: You can have the edicine ball thrown with (a) the sae speed as the baseball, (b) the sae oentu, or (c) the sae kinetic energy. Rank these choices fro easiest to hardest to catch. The Floating Astronaut A SkyLab astronaut discovered that while concentrating on writing soe notes, he had gradually floated to the iddle of an open area in the spacecraft. Not wanting to wait until he floated to the opposite side, he asked his colleagues for a push. Laughing at his predicaent, they decided not to help, and so he had to take off his unifor and throw it in one direction so that he would be propelled in the opposite direction. Estiate his resulting velocity. We begin by aking soe reasonable guesses of relevant data. Let us assue we have a 70-kg astronaut who threw his -kg unifor at a speed of 20 /s. For conve- v f Figure 9.2 soewhere. A hapless astronaut has discarded his unifor to get v 2f

5 9.2 Ipulse and Moentu CHAPTER 9 Linear Moentu and Collisions nience, we set the positive direction of the ais to be the direction of the throw (Fig. 9.2). Let us also assue that the ais is tangent to the circular path of the spacecraft. We take the syste to consist of the astronaut and the unifor. Because of the gravitational force (which keeps the astronaut, his unifor, and the entire spacecraft in orbit), the syste is not really isolated. However, this force is directed perpendicular to the otion of the syste. Therefore, oentu is constant in the direction because there are no eternal forces in this direction. The total oentu of the syste before the throw is zero ( v i 2 v 2i 0). Therefore, the total oentu after the throw ust be zero; that is, v f 2 v 2f 0 With 70 kg, v 2f 20i /s, and 2 kg, solving for v f, we find the recoil velocity of the astronaut to be v f 2 v 2f 0.3i /s kg (20i /s) 70 kg The negative sign for v f indicates that the astronaut is oving to the left after the throw, in the direction opposite the direction of otion of the unifor, in accordance with Newton s third law. Because the astronaut is uch ore assive than his unifor, his acceleration and consequent velocity are uch saller than the acceleration and velocity of the unifor. Ipulse of a force Ipulse oentu theore changes fro p i at tie t i to p f at tie t f, integrating Equation 9.7 gives p p f p i tf (9.8) To evaluate the integral, we need to know how the force varies with tie. The quantity on the right side of this equation is called the ipulse of the force F acting on a particle over the tie interval t t f t i. Ipulse is a vector defined by I tf ti F dt p The ipulse of the force F acting on a particle equals the change in the oentu of the particle caused by that force. ti F dt (9.9) 6.3 & EXAMPLE 9.2 Breakup of a Kaon at Rest ne type of nuclear particle, called the neutral kaon (K 0 ), breaks up into a pair of other particles called pions ( and ) that are oppositely charged but equal in ass, as illustrated in Figure 9.3. Assuing the kaon is initially at rest, prove that the two pions ust have oenta that are equal in agnitude and opposite in direction. The breakup of the kaon can be written K 0 9: If we let p be the oentu of the positive pion and p the oentu of the negative pion, the final oentu of the syste consisting of the two pions can be written p f p p Because the kaon is at rest before the breakup, we know that p i 0. Because oentu is conserved, p i p f 0, so that p p 0, or p p IMPULSE AND MMENTUM As we have seen, the oentu of a particle changes if a net force acts on the particle. Knowing the change in oentu caused by a force is useful in solving soe types of probles. To begin building a better understanding of this iportant concept, let us assue that a single force F acts on a particle and that this force ay vary with tie. According to Newton s second law, F dp/dt, or dp F dt (9.7) We can integrate 2 this epression to find the change in the oentu of a particle when the force acts over soe tie interval. If the oentu of the particle The iportant point behind this proble is that even though it deals with objects that are very different fro those in the preceding eaple, the physics is identical: Linear oentu is conserved in an isolated syste. Figure Note that here we are integrating force with respect to tie. Copare this with our efforts in Chapter 7, where we integrated force with respect to position to epress the work done by the force. p 0 Κ Before decay (at rest) π π + After decay A kaon at rest breaks up spontaneously into a pair of oppositely charged pions. The pions ove apart with oenta that are equal in agnitude but opposite in direction. p + F F F t i t i (a) Area = F t (b) Figure 9.4 (a) A force acting on a particle ay vary in tie. The ipulse iparted to the particle by the force is the area under the force versus tie curve. (b) In the tie interval t, the tie-averaged force (horizontal dashed line) gives the sae ipulse to a particle as does the tie-varying force described in part (a). t f t f t t This stateent, known as the ipulse oentu theore, 3 is equivalent to Newton s second law. Fro this definition, we see that ipulse is a vector quantity having a agnitude equal to the area under the force tie curve, as described in Figure 9.4a. In this figure, it is assued that the force varies in tie in the general anner shown and is nonzero in the tie interval t t f t i. The direction of the ipulse vector is the sae as the direction of the change in oentu. Ipulse has the diensions of oentu that is, ML/T. Note that ipulse is not a property of a particle; rather, it is a easure of the degree to which an eternal force changes the oentu of the particle. Therefore, when we say that an ipulse is given to a particle, we ean that oentu is transferred fro an eternal agent to that particle. Because the force iparting an ipulse can generally vary in tie, it is convenient to define a tie-averaged force F t tf F dt (9.0) where t t f t i. (This is an application of the ean value theore of calculus.) Therefore, we can epress Equation 9.9 as I F t (9.) This tie-averaged force, described in Figure 9.4b, can be thought of as the constant force that would give to the particle in the tie interval t the sae ipulse that the tie-varying force gives over this sae interval. In principle, if F is known as a function of tie, the ipulse can be calculated fro Equation 9.9. The calculation becoes especially siple if the force acting on the particle is constant. In this case, F F and Equation 9. becoes I F t (9.2) In any physical situations, we shall use what is called the ipulse approiation, in which we assue that one of the forces eerted on a particle acts for a short tie but is uch greater than any other force present. This approiation is especially useful in treating collisions in which the duration of the 3 Although we assued that only a single force acts on the particle, the ipulse oentu theore is valid when several forces act; in this case, we replace F in Equation 9.9 with F. ti

6 9.2 Ipulse and Moentu CHAPTER 9 Linear Moentu and Collisions the ball is v B C g (200 )(9.80 /s 2 ) 44 /s Considering the tie interval for the collision, v i v A 0 and v f v B for the ball. Hence, the agnitude of the ipulse iparted to the ball is I p v B v A ( kg)(44 /s) kg/s Eercise If the club is in contact with the ball for a tie of s, estiate the agnitude of the average force eerted by the club on the ball. Answer N, a value that is etreely large when copared with the weight of the ball, 0.49 N. Figure 9.5 A golf ball being struck by a club. ( Harold E. Edgerton/ Courtesy of Pal Press, Inc.) During the brief tie the club is in contact with the ball, the ball gains oentu as a result of the collision, and the club loses the sae aount of oentu. collision is very short. When this approiation is ade, we refer to the force as an ipulsive force. For eaple, when a baseball is struck with a bat, the tie of the collision is about 0.0 s and the average force that the bat eerts on the ball in this tie is typically several thousand newtons. Because this is uch greater than the agnitude of the gravitational force, the ipulse approiation justifies our ignoring the weight of the ball and bat. When we use this approiation, it is iportant to reeber that p i and p f represent the oenta iediately before and after the collision, respectively. Therefore, in any situation in which it is proper to use the ipulse approiation, the particle oves very little during the collision. Quick Quiz 9.3 Two objects are at rest on a frictionless surface. bject has a greater ass than object 2. When a force is applied to object, it accelerates through a distance d. The force is reoved fro object and is applied to object 2. At the oent when object 2 has accelerated through the sae distance d, which stateents are true? (a) p p 2, (b) p p 2, (c) p p 2, (d) K K 2, (e) K K 2, (f) K K 2. QuickLab If you can find soeone willing, play catch with an egg. What is the best way to ove your hands so that the egg does not break when you change its oentu to zero? EXAMPLE 9.4 How Good Are the Bupers? In a particular crash test, an autoobile of ass 500 kg collides with a wall, as shown in Figure 9.6. The initial and final velocities of the autoobile are v i 5.0i /s and v f 2.60i /s, respectively. If the collision lasts for 0.50 s, find the ipulse caused by the collision and the average force eerted on the autoobile. Let us assue that the force eerted on the car by the wall is large copared with other forces on the car so that we can apply the ipulse approiation. Furtherore, we note that the force of gravity and the noral force eerted by the road on the car are perpendicular to the otion and therefore do not affect the horizontal oentu. 5.0 /s The initial and final oenta of the autoobile are p i v i ( 500 kg)(5.0i /s) i kg/s p f v f ( 500 kg)(2.60i /s) i kg/s Hence, the ipulse is I p p f p i i kg/s I ( i kg/s) i kg/s The average force eerted on the autoobile is F p i kg/s t 0.50 s Before i N EXAMPLE 9.3 Teeing ff A golf ball of ass 50 g is struck with a club (Fig. 9.5). The force eerted on the ball by the club varies fro zero, at the instant before contact, up to soe aiu value (at which the ball is defored) and then back to zero when the ball leaves the club. Thus, the force tie curve is qualitatively described by Figure 9.4. Assuing that the ball travels 200, estiate the agnitude of the ipulse caused by the collision. Let us use to denote the oent when the club first contacts the ball, to denote the oent when the club loses contact with the ball as the ball starts on its trajectory, and to denote its landing. Neglecting air resistance, we can use Equation 4.4 for the range of a projectile: R C v 2 B sin 2 B g Let us assue that the launch angle B is 45, the angle that provides the aiu range for any given launch velocity. This assuption gives sin 2 B, and the launch velocity of Figure 9.6 (a) This car s oentu changes as a result of its collision with the wall. (b) In a crash test, uch of the car s initial kinetic energy is transfored into energy used to daage the car /s (a) After (b)

7 9.3 Collisions CHAPTER 9 Linear Moentu and Collisions Note that the agnitude of this force is large copared with the weight of the car ( g N), which justifies our initial assuption. f note in this proble is how the signs of the velocities indicated the reversal of directions. What would the atheatics be describing if both the initial and final velocities had the sae sign? Moentu is conserved for any collision Therefore, we conclude that the total oentu of an isolated syste just before a collision equals the total oentu of the syste just after the collision. 6.5 & 6.6 Quick Quiz 9.4 Rank an autoobile dashboard, seatbelt, and airbag in ters of (a) the ipulse and (b) the average force they deliver to a front-seat passenger during a collision. 9.3 CLLISINS In this section we use the law of conservation of linear oentu to describe what happens when two particles collide. We use the ter collision to represent the event of two particles coing together for a short tie and thereby producing ipulsive forces on each other. These forces are assued to be uch greater than any eternal forces present. A collision ay entail physical contact between two acroscopic objects, as described in Figure 9.7a, but the notion of what we ean by collision ust be generalized because physical contact on a subicroscopic scale is ill-defined and hence eaningless. To understand this, consider a collision on an atoic scale (Fig. 9.7b), such as the collision of a proton with an alpha particle (the nucleus of a heliu ato). Because the particles are both positively charged, they never coe into physical contact with each other; instead, they repel each other because of the strong electrostatic force between the at close separations. When two particles and 2 of asses and 2 collide as shown in Figure 9.7, the ipulsive forces ay vary in tie in coplicated ways, one of which is described in Figure 9.8. If F 2 is the force eerted by particle 2 on particle, and if we assue that no eternal forces act on the particles, then the change in oentu of particle due to the collision is given by Equation 9.8: p tf F 2 dt Likewise, if F 2 is the force eerted by particle on particle 2, then the change in oentu of particle 2 is p 2 tf F 2 dt Fro Newton s third law, we conclude that p p 2 p p 2 0 Because the total oentu of the syste is p syste p p 2, we conclude that the change in the oentu of the syste due to the collision is zero: p syste p p 2 constant This is precisely what we epect because no eternal forces are acting on the syste (see Section 9.2). Because the ipulsive forces are internal, they do not change the total oentu of the syste (only eternal forces can do that). ti ti F 2 F p + 2 (a) ++ 4He (b) F 2 F 2 F 2 Figure 9.7 (a) The collision between two objects as the result of direct contact. (b) The collision between two charged particles. Figure 9.8 The ipulse force as a function of tie for the two colliding particles described in Figure 9.7a. Note that F 2 F 2. t EXAMPLE 9.5 Carry Collision Insurance! A car of ass 800 kg stopped at a traffic light is struck fro the rear by a 900-kg car, and the two becoe entangled. If the saller car was oving at 20.0 /s before the collision, what is the velocity of the entangled cars after the collision? We can guess that the final speed is less than 20.0 /s, the initial speed of the saller car. The total oentu of the syste (the two cars) before the collision ust equal the total oentu iediately after the collision because oentu is conserved in any type of collision. The agnitude of the total oentu before the collision is equal to that of the saller car because the larger car is initially at rest: p i v i (900 kg)(20.0 /s) kg/s After the collision, the agnitude of the oentu of When the bowling ball and pin collide, part of the ball s oentu is transferred to the pin. Consequently, the pin acquires oentu and kinetic energy, and the ball loses oentu and kinetic energy. However, the total oentu of the syste (ball and pin) reains constant. Elastic collision Quick Quiz 9.5 As a ball falls toward the Earth, the ball s oentu increases because its speed increases. Does this ean that oentu is not conserved in this situation? Quick Quiz 9.6 A skater is using very low-friction rollerblades. A friend throws a Frisbee straight at her. In which case does the Frisbee ipart the greatest ipulse to the skater: (a) she catches the Frisbee and holds it, (b) she catches it oentarily but drops it, (c) she catches it and at once throws it back to her friend? 9.4 the entangled cars is Equating the oentu before to the oentu after and solving for v f, the final velocity of the entangled cars, we have p i v f 2 p f ( 2 )v f (2 700 kg)v f kg/s kg The direction of the final velocity is the sae as the velocity of the initially oving car. Eercise What would be the final speed if the two cars each had a ass of 900 kg? Answer 0.0 /s. ELASTIC AND INELASTIC CLLISINS IN NE DIMENSIN 6.67 /s As we have seen, oentu is conserved in any collision in which eternal forces are negligible. In contrast, kinetic energy ay or ay not be constant, depending on the type of collision. In fact, whether or not kinetic energy is the sae before and after the collision is used to classify collisions as being either elastic or inelastic. An elastic collision between two objects is one in which total kinetic energy (as well as total oentu) is the sae before and after the collision. Billiard-ball collisions and the collisions of air olecules with the walls of a container at ordinary teperatures are approiately elastic. Truly elastic collisions do occur, however, between atoic and subatoic particles. Collisions between certain objects in the acroscopic world, such as billiard-ball collisions, are only approiately elastic because soe deforation and loss of kinetic energy take place.

8 9.4 Elastic and Inelastic Collisions in ne Diension CHAPTER 9 Linear Moentu and Collisions An inelastic collision is one in which total kinetic energy is not the sae before and after the collision (even though oentu is constant). Inelastic collisions are of two types. When the colliding objects stick together after the collision, as happens when a eteorite collides with the Earth, the collision is called perfectly inelastic. When the colliding objects do not stick together, but soe kinetic energy is lost, as in the case of a rubber ball colliding with a hard surface, the collision is called inelastic (with no odifying adverb). For eaple, when a rubber ball collides with a hard surface, the collision is inelastic because soe of the kinetic energy of the ball is lost when the ball is defored while it is in contact with the surface. In ost collisions, kinetic energy is not the sae before and after the collision because soe of it is converted to internal energy, to elastic potential energy when the objects are defored, and to rotational energy. Elastic and perfectly inelastic collisions are liiting cases; ost collisions fall soewhere between the. In the reainder of this section, we treat collisions in one diension and consider the two etree cases perfectly inelastic and elastic collisions. The iportant distinction between these two types of collisions is that oentu is constant in all collisions, but kinetic energy is constant only in elastic collisions. Inelastic collision QuickLab Hold a Ping-Pong ball or tennis ball on top of a basketball. Drop the both at the sae tie so that the basketball hits the floor, bounces up, and hits the saller falling ball. What happens and why? Before collision v i v 2i 2 v f (a) After collision v 2f (b) Figure 9.0 Scheatic representation of an elastic head-on collision between two particles: (a) before collision and (b) after collision. if it oves to the left. As has been seen in earlier chapters, it is coon practice to call these values speed even though this ter technically refers to the agnitude of the velocity vector, which does not have an algebraic sign. In a typical proble involving elastic collisions, there are two unknown quantities, and Equations 9.5 and 9.6 can be solved siultaneously to find these. An alternative approach, however one that involves a little atheatical anipulation of Equation 9.6 often siplifies this process. To see how, let us cancel the factor 2 in Equation 9.6 and rewrite it as (v 2 i v 2 f ) 2 (v 2 2f v 2 2i ) and then factor both sides: (v i v f )(v i v f ) 2 (v 2f v 2i )(v 2f v 2i ) (9.7) Net, let us separate the ters containing and 2 in Equation 9.5 to get (v i v f ) 2 (v 2f v 2i ) (9.8) To obtain our final result, we divide Equation 9.7 by Equation 9.8 and get v i v f v 2f v 2i 6.6 Perfectly Inelastic Collisions Consider two particles of asses and 2 oving with initial velocities v i and v 2i along a straight line, as shown in Figure 9.9. The two particles collide head-on, stick together, and then ove with soe coon velocity v f after the collision. Because oentu is conserved in any collision, we can say that the total oentu before the collision equals the total oentu of the coposite syste after the collision: v i 2 v 2i ( 2 )v f (9.3) Quick Quiz 9.7 v f v i 2 v 2i 2 (9.4) Which is worse, crashing into a brick wall at 40 i/h or crashing head-on into an oncoing car that is identical to yours and also oving at 40 i/h? Elastic Collisions Now consider two particles that undergo an elastic head-on collision (Fig. 9.0). In this case, both oentu and kinetic energy are conserved; therefore, we have v i 2 v 2i v f 2 v 2f (9.5) Before collision 2 v i v 2i (a) After collision + 2 (b) Figure 9.9 Scheatic representation of a perfectly inelastic head-on collision between two particles: (a) before collision and (b) after collision. v f Elastic collision: relationships between final and initial velocities Elastic collision: particle 2 initially at rest v i v 2i (v f v 2f ) (9.9) This equation, in cobination with Equation 9.5, can be used to solve probles dealing with elastic collisions. According to Equation 9.9, the relative speed of the two particles before the collision v i v 2i equals the negative of their relative speed after the collision, (v f v 2f ). Suppose that the asses and initial velocities of both particles are known. Equations 9.5 and 9.9 can be solved for the final speeds in ters of the initial speeds because there are two equations and two unknowns: v f 2 2 v i v 2i v 2f 2 2 v i 2 2 v 2i (9.20) (9.2) It is iportant to reeber that the appropriate signs for v i and v 2i ust be included in Equations 9.20 and 9.2. For eaple, if particle 2 is oving to the left initially, then v 2i is negative. Let us consider soe special cases: If 2, then v f v 2i and v 2f v i. That is, the particles echange speeds if they have equal asses. This is approiately what one observes in head-on billiard ball collisions the cue ball stops, and the struck ball oves away fro the collision with the sae speed that the cue ball had. If particle 2 is initially at rest, then v 2i 0 and Equations 9.20 and 9.2 becoe v f 2 2 v i v 2f 2 2 v i (9.22) (9.23) 2 v 2 i 2 2v 2 2i 2 v 2 f 2 2 2v 2f (9.6) Because all velocities in Figure 9.0 are either to the left or the right, they can be represented by the corresponding speeds along with algebraic signs indicating direction. We shall indicate v as positive if a particle oves to the right and negative If is uch greater than 2 and v 2i 0, we see fro Equations 9.22 and 9.23 that v f v i and v 2f 2v i. That is, when a very heavy particle collides headon with a very light one that is initially at rest, the heavy particle continues its o-

9 9.4 Elastic and Inelastic Collisions in ne Diension CHAPTER 9 Linear Moentu and Collisions tion unaltered after the collision, and the light particle rebounds with a speed equal to about twice the initial speed of the heavy particle. An eaple of such a collision would be that of a oving heavy ato, such as uraniu, with a light ato, such as hydrogen. If 2 is uch greater than and particle 2 is initially at rest, then v f v i and v 2f v 2i 0. That is, when a very light particle collides head-on with a very heavy particle that is initially at rest, the light particle has its velocity reversed and the heavy one reains approiately at rest. EXAMPLE 9.6 The Ballistic Pendulu The ballistic pendulu (Fig. 9.) is a syste used to easure the speed of a fast-oving projectile, such as a bullet. The bullet is fired into a large block of wood suspended fro soe light wires. The bullet ebeds in the block, and the entire syste swings through a height h. The collision is perfectly inelastic, and because oentu is conserved, Equation 9.4 gives the speed of the syste right after the collision, when we assue the ipulse approiation. If we call the bullet particle and the block particle 2, the total kinetic energy right after the collision is () K f 2 ( 2 )v 2 f With v 2i 0, Equation 9.4 becoes Eercise In a ballistic pendulu eperient, suppose that h 5.00 c, 5.00 g, and 2.00 kg. Find (a) the initial speed of the bullet and (b) the loss in echanical energy due to the collision. Answer 99 /s; 98.5 J. EXAMPLE 9.7 A Two-Body Collision with a Spring A block of ass.60 kg initially oving to the right with a speed of 4.00 /s on a frictionless horizontal track collides with a spring attached to a second block of ass kg initially oving to the left with a speed of 2.50 /s, as shown in Figure 9.2a. The spring constant is 600 N/. (a) At the instant block is oving to the right with a speed of 3.00 /s, as in Figure 9.2b, deterine the velocity of block 2. First, note that the initial velocity of block 2 is 2.50 /s because its direction is to the left. Because oentu is conserved for the syste of two blocks, we have v i 2 v 2i v f 2 v 2f (.60 kg)(4.00 /s) (2.0 kg)(2.50 /s) v 2f (.60 kg)(3.00 /s) (2.0 kg)v 2f.74 /s The negative value for v 2f eans that block 2 is still oving to the left at the instant we are considering. (b) Deterine the distance the spring is copressed at that instant. To deterine the distance that the spring is copressed, shown as in Figure 9.2b, we can use the concept of conservation of echanical energy because no friction or other nonconservative forces are acting on the syste. Thus, we have 2 v 2 i 2 2v 2 2i 2 v 2 f 2 2v 2 2f 2 k2 Substituting the given values and the result to part (a) into this epression gives 0.73 It is iportant to note that we needed to use the principles of both conservation of oentu and conservation of echanical energy to solve the two parts of this proble. Eercise Find the velocity of block and the copression in the spring at the instant that block 2 is at rest. Answer 0.79 /s to the right; (2) v f v i 2 Substituting this value of v f into () gives Note that this kinetic energy iediately after the collision is less than the initial kinetic energy of the bullet. In all the energy changes that take place after the collision, however, the total aount of echanical energy reains constant; thus, we can say that after the collision, the kinetic energy of the block and bullet at the botto is transfored to potential energy at the height h: Solving for v i, we obtain K f 2 v 2 i 2( 2 ) 2 v 2 i 2( 2 ) ( 2 )gh v i 2 2gh This epression tells us that it is possible to obtain the initial speed of the bullet by easuring h and the two asses. Because the collision is perfectly inelastic, soe echanical energy is converted to internal energy and it would be incorrect to equate the initial kinetic energy of the incoing bullet to the final gravitational potential energy of the bullet block cobination. v i v f 2 Figure 9. (a) (b) + 2 (a) Diagra of a ballistic pendulu. Note that v i is the velocity of the bullet just before the collision and v f v f v 2f is the velocity of the bullet block syste just after the perfectly inelastic collision. (b) Multiflash photograph of a ballistic pendulu used in the laboratory. h v i = (4.00i) /s 2 (a) k v 2i = ( 2.50i) /s Figure 9.2 v f = (3.00i) /s k 2 EXAMPLE 9.8 Slowing Down Neutrons by Collisions In a nuclear reactor, neutrons are produced when a 92U Let us assue that the oderator nucleus of ato splits in a process called fission. These neutrons are oving at about 0 7 /s and ust be slowed down to about 0 3 /s before they take part in another fission event. They are slowed down by being passed through a solid or liquid aterial called a oderator. The slowing-down process involves elastic collisions. Let us show that a neutron can lose ost of its kinetic energy if it collides elastically with a oderator containing light nuclei, such as deuteriu (in heavy water, D 2 ) or carbon (in graphite). ass is at rest initially and that a neutron of ass n and initial speed v ni collides with it head-on. Because these are elastic collisions, the first thing we do is recognize that both oentu and kinetic energy are constant. Therefore, Equations 9.22 and 9.23 can be applied to the head-on collision of a neutron with a oderator nucleus. We can represent this process by a drawing such as Figure 9.0. The initial kinetic energy of the neutron is (b) v 2f

10 9.4 Elastic and Inelastic Collisions in ne Diension CHAPTER 9 Linear Moentu and Collisions After the collision, the neutron has kinetic energy 2 nv 2 nf, and we can substitute into this the value for v nf given by Equation 9.22: Therefore, the fraction f n of the initial kinetic energy possessed by the neutron after the collision is () Fro this result, we see that the final kinetic energy of the neutron is sall when is close to n and zero when n. We can use Equation 9.23, which gives the final speed of the particle that was initially at rest, to calculate the kinetic energy of the oderator nucleus after the collision: Quick Quiz 9.8 K ni 2 2 nv ni K nf 2 nv 2 nf n 2 n n 2 v 2 ni f n K nf K ni n n 2 An ingenious device that illustrates conservation of oentu and kinetic energy is shown in Figure 9.3a. It consists of five identical hard balls supported by strings of equal lengths. When ball is pulled out and released, after the alost-elastic collision between it and ball 2, ball 5 oves out, as shown in Figure 9.3b. If balls and 2 are pulled out and released, balls 4 and 5 swing out, and so forth. Is it ever possible that, when ball is released, balls 4 and 5 will swing out on the opposite side and travel with half the speed of ball, as in Figure 9.3c? v K f 2 v 2 f 2 n 2 ( n ) 2 v 2 ni Hence, the fraction f of the initial kinetic energy transferred to the oderator nucleus is (2) f K f 4 n K ni ( n ) 2 Because the total kinetic energy of the syste is conserved, (2) can also be obtained fro () with the condition that f n f, so that f f n. Suppose that heavy water is used for the oderator. For collisions of the neutrons with deuteriu nuclei in D 2 ( 2 n ), f n /9 and f 8/9. That is, 89% of the neutron s kinetic energy is transferred to the deuteriu nucleus. In practice, the oderator efficiency is reduced because head-on collisions are very unlikely. How do the results differ when graphite ( 2 C, as found in pencil lead) is used as the oderator? This can happen. (b) v 9.5 TW-DIMENSINAL CLLISINS In Sections 9. and 9.3, we showed that the oentu of a syste of two particles is constant when the syste is isolated. For any collision of two particles, this result iplies that the oentu in each of the directions, y, and z is constant. However, an iportant subset of collisions takes place in a plane. The gae of billiards is a failiar eaple involving ultiple collisions of objects oving on a twodiensional surface. For such two-diensional collisions, we obtain two coponent equations for conservation of oentu: v i 2 v 2i v f 2 v 2 f v iy 2 v 2iy v fy 2 v 2fy Let us consider a two-diensional proble in which particle of ass collides with particle 2 of ass 2, where particle 2 is initially at rest, as shown in Figure 9.4. After the collision, particle oves at an angle with respect to the horizontal and particle 2 oves at an angle with respect to the horizontal. This is called a glancing collision. Applying the law of conservation of oentu in coponent for, and noting that the initial y coponent of the oentu of the two-particle syste is zero, we obtain v i v f cos 2 v 2f cos (9.24) 0 v f sin 2 v 2f sin (9.25) where the inus sign in Equation 9.25 coes fro the fact that after the collision, particle 2 has a y coponent of velocity that is downward. We now have two independent equations. As long as no ore than two of the seven quantities in Equations 9.24 and 9.25 are unknown, we can solve the proble. If the collision is elastic, we can also use Equation 9.6 (conservation of kinetic energy), with v 2i 0, to give 2 v 2 i 2 v 2 f 2 2 2v 2f (9.26) Knowing the initial speed of particle and both asses, we are left with four unknowns (v f, v 2f,, ). Because we have only three equations, one of the four reaining quantities ust be given if we are to deterine the otion after the collision fro conservation principles alone. If the collision is inelastic, kinetic energy is not conserved and Equation 9.26 does not apply. v f sin θ v f (a) Figure v Can this happen? (c) An eecutive stress reliever. 4 5 v/2 v f cos θ v i θ φ v 2f cos φ v 2f sin φ v 2f (a) Before the collision (b) After the collision Figure 9.4 An elastic glancing collision between two particles.

11 9.5 Two-Diensional Collisions CHAPTER 9 Linear Moentu and Collisions Proble-Solving Hints Collisions The following procedure is recoended when dealing with probles involving collisions between two objects: Set up a coordinate syste and define your velocities with respect to that syste. It is usually convenient to have the ais coincide with one of the initial velocities. In your sketch of the coordinate syste, draw and label all velocity vectors and include all the given inforation. Write epressions for the and y coponents of the oentu of each object before and after the collision. Reeber to include the appropriate signs for the coponents of the velocity vectors. Write epressions for the total oentu in the direction before and after the collision and equate the two. Repeat this procedure for the total oentu in the y direction. These steps follow fro the fact that, because the oentu of the syste is conserved in any collision, the total oentu along any direction ust also be constant. Reeber, it is the oentu of the syste that is constant, not the oenta of the individual objects. If the collision is inelastic, kinetic energy is not conserved, and additional inforation is probably required. If the collision is perfectly inelastic, the final velocities of the two objects are equal. Solve the oentu equations for the unknown quantities. If the collision is elastic, kinetic energy is conserved, and you can equate the total kinetic energy before the collision to the total kinetic energy after the collision to get an additional relationship between the velocities. EXAMPLE 9.9 Collision at an Intersection A 500-kg car traveling east with a speed of 25.0 /s collides at an intersection with a kg van traveling north at a speed of 20.0 /s, as shown in Figure 9.5. Find the direction and agnitude of the velocity of the wreckage after the collision, assuing that the vehicles undergo a perfectly inelastic collision (that is, they stick together). Let us choose east to be along the positive direction and north to be along the positive y direction. Before the collision, the only object having oentu in the direction is the car. Thus, the agnitude of the total initial oentu of the syste (car plus van) in the direction is p i ( 500 kg)(25.0 /s) kg/s Let us assue that the wreckage oves at an angle and speed v f after the collision. The agnitude of the total oentu in the direction after the collision is p f (4 000 kg)v f cos Because the total oentu in the direction is constant, we can equate these two equations to obtain () kg/s (4 000 kg)v f cos Figure 9.5 (25.0i) /s y Siilarly, the total initial oentu of the syste in the y direction is that of the van, and the agnitude of this oentu is (2 500 kg)(20.0 /s). Applying conservation of θ v f (20.0j) /s An eastbound car colliding with a northbound van. oentu to the y direction, we have (2) kg/s (4 000 kg)v f sin If we divide (2) by (), we get sin tan cos EXAMPLE 9.0 Proton Proton Collision Proton collides elastically with proton 2 that is initially at rest. Proton has an initial speed of /s and akes a glancing collision with proton 2, as was shown in Figure 9.4. After the collision, proton oves at an angle of 37.0 to the horizontal ais, and proton 2 deflects at an angle to the sae ais. Find the final speeds of the two protons and the angle. Because both particles are protons, we know that 2. We also know that 37.0 and v i /s. Equations 9.24, 9.25, and 9.26 becoe v f cos 37.0 v 2f cos /s v f sin 37.0 v 2f sin 0 EXAMPLE 9. p yi p yf (2 500 kg)(20.0 /s) (4 000 kg)v f sin v 2 f v 2 2f ( /s) 2 Billiard Ball Collision In a gae of billiards, a player wishes to sink a target ball 2 in the corner pocket, as shown in Figure 9.6. If the angle to the corner pocket is 35, at what angle is the cue ball deflected? Assue that friction and rotational otion are uniportant and that the collision is elastic. Because the target ball is initially at rest, conservation of energy (Eq. 9.6) gives 2 v 2 i 2 v 2 f 2 2 2v 2f But 2, so that () v 2 i v 2 f v 2 2f Applying conservation of oentu to the two-diensional collision gives (2) v i v f v 2f Note that because 2, the asses also cancel in (2). If we square both sides of (2) and use the definition of the dot When this angle is substituted into (2), the value of v f is v f kg/s (4 000 kg)sin 53. It ight be instructive for you to draw the oentu vectors of each vehicle before the collision and the two vehicles together after the collision. Solving these three equations with three unknowns siultaneously gives v f v 2f /s /s /s Note that 90. This result is not accidental. Whenever two equal asses collide elastically in a glancing collision and one of the is initially at rest, their final velocities are always at right angles to each other. The net eaple illustrates this point in ore detail. v i Cue ball y θ 35 Figure 9.6 v 2f v f

12 9.6 The Center of Mass CHAPTER 9 Linear Moentu and Collisions product of two vectors fro Section 7.2, we get v i 2 (v f v 2f ) (v f v 2f ) v f 2 v 2f 2 2v f v 2f Because the angle between v f and v 2f is 35, v f v 2f v f v 2f cos( 35 ), and so (3) v 2 i v 2 f v 2 2f 2v f v 2f cos( 35 ) Subtracting () fro (3) gives 0 2v f v 2f cos( 35 ) THE CENTER FMASS In this section we describe the overall otion of a echanical syste in ters of a special point called the center of ass of the syste. The echanical syste can be either a syste of particles, such as a collection of atos in a container, or an etended object, such as a gynast leaping through the air. We shall see that the center of ass of the syste oves as if all the ass of the syste were concentrated at that point. Furtherore, if the resultant eternal force on the syste is F et and the total ass of the syste is M, the center of ass oves with an acceleration given by a F et /M. That is, the syste oves as if the resultant eternal force were applied to a single particle of ass M located at the center of ass. This behavior is independent of other otion, such as rotation or vibration of the syste. This result was iplicitly assued in earlier chapters because any eaples referred to the otion of etended objects that were treated as particles. Consider a echanical syste consisting of a pair of particles that have different asses and are connected by a light, rigid rod (Fig. 9.7). ne can describe the position of the center of ass of a syste as being the average position of the syste s ass. The center of ass of the syste is located soewhere on the line joining the This ultiflash photograph shows that as the acrobat eecutes a soersault, his center of ass follows a parabolic path, the sae path that a particle would follow. 0 cos( 35 ) or 55 This result shows that whenever two equal asses undergo a glancing elastic collision and one of the is initially at rest, they ove at right angles to each other after the collision. The sae physics describes two very different situations, protons in Eaple 9.0 and billiard balls in this eaple. CM CM CM (a) (b) (c) Figure 9.7 Two particles of unequal ass are connected by a light, rigid rod. (a) The syste rotates clockwise when a force is applied between the less assive particle and the center of ass. (b) The syste rotates counterclockwise when a force is applied between the ore assive particle and the center of ass. (c) The syste oves in the direction of the force without rotating when a force is applied at the center of ass. y 2 Figure 9.8 The center of ass of two particles of unequal ass on the ais is located at CM, a point between the particles, closer to the one having the larger ass. Vector position of the center of ass for a syste of particles y CM r i r CM i CM CM 2 z Figure 9.9 An etended object can be considered a distribution of sall eleents of ass i. The center of ass is located at the vector position r CM, which has coordinates CM, y CM, and z CM. particles and is closer to the particle having the larger ass. If a single force is applied at soe point on the rod soewhere between the center of ass and the less assive particle, the syste rotates clockwise (see Fig. 9.7a). If the force is applied at a point on the rod soewhere between the center of ass and the ore assive particle, the syste rotates counterclockwise (see Fig. 9.7b). If the force is applied at the center of ass, the syste oves in the direction of F without rotating (see Fig. 9.7c). Thus, the center of ass can be easily located. The center of ass of the pair of particles described in Figure 9.8 is located on the ais and lies soewhere between the particles. Its coordinate is CM 2 2 (9.27) 2 For eaple, if and we find that CM 2 0, 2 d, 2 2, 3 d. That is, the center of ass lies closer to the ore assive particle. If the two asses are equal, the center of ass lies idway between the particles. We can etend this concept to a syste of any particles in three diensions. The coordinate of the center of ass of n particles is defined to be CM n i i n i (9.28) 2 3 n i i where i is the coordinate of the ith particle. For convenience, we epress the total ass as M i, where the su runs over all n particles. The y and z coordinates of the center of ass are siilarly defined by the equations i i y i i z i y CM i and z CM i (9.29) M M The center of ass can also be located by its position vector, r CM. The cartesian coordinates of this vector are CM, y CM, and z CM, defined in Equations 9.28 and Therefore, r CM CM i y CM j z CM k i i i i y i j i z i k i i i M i r i r CM i (9.30) M where r i is the position vector of the ith particle, defined by r i i i y i j z i k Although locating the center of ass for an etended object is soewhat ore cubersoe than locating the center of ass of a syste of particles, the basic ideas we have discussed still apply. We can think of an etended object as a syste containing a large nuber of particles (Fig. 9.9). The particle separation is very sall, and so the object can be considered to have a continuous ass distribution. By dividing the object into eleents of ass i, with coordinates i, y i, z i, we see that the coordinate of the center of ass is approiately i i CM i M with siilar epressions for y CM and z CM. If we let the nuber of eleents n approach infinity, then CM is given precisely. In this liit, we replace the su by an

13 9.6 The Center of Mass CHAPTER 9 Linear Moentu and Collisions integral and i by the differential eleent d: i i CM li i i:0 M d M Likewise, for y CM and z CM we obtain (9.3) (9.32) We can epress the vector position of the center of ass of an etended object in the for (9.33) which is equivalent to the three epressions given by Equations 9.3 and The center of ass of any syetric object lies on an ais of syetry and on any plane of syetry. 4 For eaple, the center of ass of a rod lies in the rod, idway between its ends. The center of ass of a sphere or a cube lies at its geoetric center. ne can deterine the center of ass of an irregularly shaped object by suspending the object first fro one point and then fro another. In Figure 9.20, a wrench is hung fro point A, and a vertical line AB (which can be established with a plub bob) is drawn when the wrench has stopped swinging. The wrench is then hung fro point C, and a second vertical line CD is drawn. The center of ass is halfway through the thickness of the wrench, under the intersection of these two lines. In general, if the wrench is hung freely fro any point, the vertical line through this point ust pass through the center of ass. Because an etended object is a continuous distribution of ass, each sall ass eleent is acted upon by the force of gravity. The net effect of all these forces is equivalent to the effect of a single force, Mg, acting through a special point, called the center of gravity. If g is constant over the ass distribution, then the center of gravity coincides with the center of ass. If an etended object is pivoted at its center of gravity, it balances in any orientation. Quick Quiz 9.9 y CM M y d and z CM M z d r CM M r d If a baseball bat is cut at the location of its center of ass as shown in Figure 9.2, do the two pieces have the sae ass? Figure 9.2 A baseball bat cut at the location of its center of ass. 4 This stateent is valid only for objects that have a unifor ass per unit volue. A B Center of ass Figure 9.20 An eperiental technique for deterining the center of ass of a wrench. The wrench is hung freely first fro point A and then fro point C. The intersection of the two lines AB and CD locates the center of ass. QuickLab Cut a triangle fro a piece of cardboard and draw a set of adjacent strips inside it, parallel to one of the sides. Put a dot at the approiate location of the center of ass of each strip and then draw a straight line through the dots and into the angle opposite your starting side. The center of ass for the triangle ust lie on this bisector of the angle. Repeat these steps for the other two sides. The three angle bisectors you have drawn will intersect at the center of ass of the triangle. If you poke a hole anywhere in the triangle and hang the cardboard fro a string attached at that hole, the center of ass will be vertically aligned with the hole. A D C B C EXAMPLE 9.2 The Center of Mass of Three Particles A syste consists of three particles located as shown in Figure 9.22a. Find the center of ass of the syste. We set up the proble by labeling the asses of the particles as shown in the figure, with 2.0 kg and kg. Using the basic defining equations for the coordinates of the center of ass and noting that z CM 0, we obtain i i CM i M (.0 kg)(.0 ) (.0 kg)(2.0 ) (2.0 kg)(0 ).0 kg.0 kg 2.0 kg 3.0 kg kg i y i y CM i M y 2 y 2 3 y (.0 kg)(0) (.0 kg)(0) (2.0 kg)(2.0 ) 4.0 kg 4.0 kg kg The position vector to the center of ass easured fro the origin is therefore r CM CM i y CM j 0.75i.0 j We can verify this result graphically by adding together r 2 r 2 3 r 3 and dividing the vector su by M, the total ass. This is shown in Figure 9.22b. Figure 9.22 (a) Two -kg asses and a single 2-kg ass are located as shown. The vector indicates the location of the syste s center of ass. (b) The vector su of i r i. EXAMPLE 9.3 The Center of Mass of a Rod (a) Show that the center of ass of a rod of ass M and length L lies idway between its ends, assuing the rod has a unifor ass per unit length. The rod is shown aligned along the ais in Figure 9.23, so that y CM z CM 0. Furtherore, if we call the ass per unit length (this quantity is called the linear ass density), then M/L for the unifor rod we assue here. If we divide the rod into eleents of length d, then the ass of each eleent is d d. For an arbitrary eleent located a distance fro the origin, Equation 9.3 gives CM M d M L 0 d M 2 2 L L 2 0 2M Mr CM Because this reduces to ne can also use syetry arguents to obtain the sae result. (b) Suppose a rod is nonunifor such that its ass per unit length varies linearly with according to the epression, where is a constant. Find the coordinate of the center of ass as a fraction of L. y() M/L, CM r CM 3 r 3 2 L2 2M M L In this case, we replace d by d where is not constant. Therefore, CM is (a) r 2 r 2 (b) L 2 3 () r CM

14 9.7 Motion of a Syste of Particles CHAPTER 9 Linear Moentu and Collisions We can eliinate by noting that the total ass of the rod is related to through the relationship Substituting this into the epression for CM gives EXAMPLE 9.4 The Center of Mass of a Right Triangle An object of ass M is in the shape of a right triangle whose With this substitution, CM becoes diensions are shown in Figure Locate the coordinates CM 2 a ab 0 b of the center of ass, assuing the object has a unifor ass a d 2 a 2 a 2 d 2 0 a 23 3 a per unit area. 0 CM M d M L 0 L L 3 2 d M 0 3M M d L 0 By inspection we can estiate that the coordinate of the center of ass ust be past the center of the base, that is, greater than a/2, because the largest part of the triangle lies beyond that point. A siilar arguent indicates that its y coordinate ust be less than b/2. To evaluate the coordinate, we divide the triangle into narrow strips of width d and height y as in Figure The ass d of each strip is d d L L 3 CM 3L 2 /2 total ass of object area of strip total area of object M /2ab (y d) 2M ab y d Therefore, the coordinate of the center of ass is CM d a M M 0 2M d M L L 2 d L d 0 ab y d 2 ab a y d 0 To evaluate this integral, we ust epress y in ters of. Fro siilar triangles in Figure 9.24, we see that y b or y b a a Figure 9.23 y 2 3 a By a siilar calculation, we get for the y coordinate of the center of ass y CM 3 b These values fit our original estiates. L d = λd λ d The center of ass of a unifor rod of length L is located at CM L/2. y c a d y Figure 9.24 d b Velocity of the center of ass Total oentu of a syste of particles Acceleration of the center of ass Newton s second law for a syste of particles definition a velocity. Assuing M reains constant for a syste of particles, that is, no particles enter or leave the syste, we get the following epression for the velocity of the center of ass of the syste: v CM dr CM dr i v i i i i (9.34) dt M i dt M where v i is the velocity of the ith particle. Rearranging Equation 9.34 gives Mv CM i i v i i p i p tot (9.35) Therefore, we conclude that the total linear oentu of the syste equals the total ass ultiplied by the velocity of the center of ass. In other words, the total linear oentu of the syste is equal to that of a single particle of ass M oving with a velocity v CM. If we now differentiate Equation 9.34 with respect to tie, we get the acceleration of the center of ass of the syste: a CM dv CM dv i i i a i dt M i dt M i Rearranging this epression and using Newton s second law, we obtain Ma CM i i a i i F i (9.36) (9.37) where F i is the net force on particle i. The forces on any particle in the syste ay include both eternal forces (fro outside the syste) and internal forces (fro within the syste). However, by Newton s third law, the internal force eerted by particle on particle 2, for eaple, is equal in agnitude and opposite in direction to the internal force eerted by particle 2 on particle. Thus, when we su over all internal forces in Equation 9.37, they cancel in pairs and the net force on the syste is caused only by eternal forces. Thus, we can write Equation 9.37 in the for F et Ma CM dp tot dt (9.38) That is, the resultant eternal force on a syste of particles equals the total ass of the syste ultiplied by the acceleration of the center of ass. If we copare this with Newton s second law for a single particle, we see that The center of ass of a syste of particles of cobined ass M oves like an equivalent particle of ass M would ove under the influence of the resultant eternal force on the syste MTIN FA SYSTEM FPARTICLES We can begin to understand the physical significance and utility of the center of ass concept by taking the tie derivative of the position vector given by Equation Fro Section 4. we know that the tie derivative of a position vector is by Finally, we see that if the resultant eternal force is zero, then fro Equation 9.38 it follows that dp tot Ma dt CM 0

15 9.7 Motion of a Syste of Particles CHAPTER 9 Linear Moentu and Collisions p CM b Figure 9.25 Multiflash photograph showing an overhead view of a wrench oving on a horizontal surface. The center of ass of the wrench oves in a straight line as the wrench rotates about this point, shown by the white dots. so that p tot Mv CM constant (when F et 0) (9.39) That is, the total linear oentu of a syste of particles is conserved if no net eternal force is acting on the syste. It follows that for an isolated syste of particles, both the total oentu and the velocity of the center of ass are constant in tie, as shown in Figure This is a generalization to a any-particle syste of the law of conservation of oentu discussed in Section 9. for a two-particle syste. Suppose an isolated syste consisting of two or ore ebers is at rest. The center of ass of such a syste reains at rest unless acted upon by an eternal force. For eaple, consider a syste ade up of a swier standing on a raft, with the syste initially at rest. When the swier dives horizontally off the raft, the center of ass of the syste reains at rest (if we neglect friction between raft and water). Furtherore, the linear oentu of the diver is equal in agnitude to that of the raft but opposite in direction. As another eaple, suppose an unstable ato initially at rest suddenly breaks up into two fragents of asses M A and M B, with velocities v A and v B, respectively. Because the total oentu of the syste before the breakup is zero, the total oentu of the syste after the breakup ust also be zero. Therefore, M A v A M B v B 0. If the velocity of one of the fragents is known, the recoil velocity of the other fragent can be calculated. Figure 9.26 The center of ass of an isolated syste reains at rest unless acted on by an eternal force. How can you deterine the ass of the polar bear? CNCEPTUAL EXAMPLE 9.6 Eploding Projectile A projectile fired into the air suddenly eplodes into several fragents (Fig. 9.27). What can be said about the otion of Motion of center of ass the center of ass of the syste ade up of all the fragents after the eplosion? Neglecting air resistance, the only eternal force on the projectile is the gravitational force. Thus, if the projectile did not eplode, it would continue to ove along the parabolic path indicated by the broken line in Figure Because the forces caused by the eplosion are internal, they do not affect the otion of the center of ass. Thus, after the eplosion the center of ass of the syste (the fragents) follows the sae parabolic path the projectile would have followed if there had been no eplosion. Figure 9.27 When a projectile eplodes into several fragents, the center of ass of the syste ade up of all the fragents follows the sae parabolic path the projectile would have taken had there been no eplosion. EXAMPLE 9.5 The Sliding Bear Suppose you tranquilize a polar bear on a sooth glacier as part of a research effort. How ight you estiate the bear s ass using a easuring tape, a rope, and knowledge of your own ass? Tie one end of the rope around the bear, and then lay out the tape easure on the ice with one end at the bear s original position, as shown in Figure Grab hold of the free end of the rope and position yourself as shown, noting your location. Take off your spiked shoes and pull on the rope hand over hand. Both you and the bear will slide over the ice until you eet. Fro the tape, observe how far you have slid, p, and how far the bear has slid, b. The point where you eet the bear is the constant location of the center of ass of the syste (bear plus you), and so you can deterine the ass of the bear fro b b p p. (Unfortunately, you cannot get back to your spiked shoes and so are in big trouble if the bear wakes up!) EXAMPLE 9.7 The Eploding Rocket A rocket is fired vertically upward. At the instant it reaches an altitude of 000 and a speed of 300 /s, it eplodes into three equal fragents. ne fragent continues to ove upward with a speed of 450 /s following the eplosion. The second fragent has a speed of 240 /s and is oving east right after the eplosion. What is the velocity of the third fragent right after the eplosion? Let us call the total ass of the rocket M; hence, the ass of each fragent is M/3. Because the forces of the eplosion are internal to the syste and cannot affect its total oentu, the total oentu p i of the rocket just before the eplosion ust equal the total oentu p f of the fragents right after the eplosion.

16 9.8 Rocket Propulsion CHAPTER 9 Linear Moentu and Collisions ptional Section 9.8 Before the eplosion: After the eplosion: p f M 3 (240i) /s M 3 (450j) /s M 3 v f where v f is the unknown velocity of the third fragent. Equating these two epressions (because p i p f ) gives M 3 v f M(80i) /s M(50j) /s M(300 j) /s RCKET p i Mv i M(300j) /s PRPULSIN When ordinary vehicles, such as autoobiles and locootives, are propelled, the driving force for the otion is friction. In the case of the autoobile, the driving force is the force eerted by the road on the car. A locootive pushes against the tracks; hence, the driving force is the force eerted by the tracks on the locootive. However, a rocket oving in space has no road or tracks to push against. Therefore, the source of the propulsion of a rocket ust be soething other than friction. Figure 9.28 is a draatic photograph of a spacecraft at liftoff. The operation of a rocket depends upon the law of conservation of linear oentu as applied to a syste of particles, where the syste is the rocket plus its ejected fuel. Rocket propulsion can be understood by first considering the echanical syste consisting of a achine gun ounted on a cart on wheels. As the gun is fired, What does the su of the oentu vectors for all the fragents look like? Eercise Find the position of the center of ass of the syste of fragents relative to the ground 3.00 s after the eplosion. Assue the rocket engine is nonoperative after the eplosion. Answer v f (240i 450j) /s The coordinate does not change; y CM.86 k. Figure 9.28 Liftoff of the space shuttle Colubia. Enorous thrust is generated by the shuttle s liquid-fuel engines, aided by the two solid-fuel boosters. Many physical principles fro echanics, therodynaics, and electricity and agnetis are involved in such a launch. The force fro a nitrogen-propelled, hand-controlled device allows an astronaut to ove about freely in space without restrictive tethers. M + p i = (M + )v (a) (b) Figure 9.29 Rocket propulsion. (a) The initial ass of the rocket plus all its fuel is M at a tie t, and its speed is v. (b) At a tie t t, the rocket s ass has been reduced to M and an aount of fuel has been ejected. The rocket s speed increases by an aount v. M Epression for rocket propulsion v v + v each bullet receives a oentu v in soe direction, where v is easured with respect to a stationary Earth frae. The oentu of the syste ade up of cart, gun, and bullets ust be conserved. Hence, for each bullet fired, the gun and cart ust receive a copensating oentu in the opposite direction. That is, the reaction force eerted by the bullet on the gun accelerates the cart and gun, and the cart oves in the direction opposite that of the bullets. If n is the nuber of bullets fired each second, then the average force eerted on the gun is F av nv. In a siilar anner, as a rocket oves in free space, its linear oentu changes when soe of its ass is released in the for of ejected gases. Because the gases are given oentu when they are ejected out of the engine, the rocket receives a copensating oentu in the opposite direction. Therefore, the rocket is accelerated as a result of the push, or thrust, fro the ehaust gases. In free space, the center of ass of the syste (rocket plus epelled gases) oves uniforly, independent of the propulsion process. 5 Suppose that at soe tie t, the agnitude of the oentu of a rocket plus its fuel is (M )v, where v is the speed of the rocket relative to the Earth (Fig. 9.29a). ver a short tie interval t, the rocket ejects fuel of ass, and so at the end of this interval the rocket s speed is v v, where v is the change in speed of the rocket (Fig. 9.29b). If the fuel is ejected with a speed v e relative to the rocket (the subscript e stands for ehaust, and v e is usually called the ehaust speed), the velocity of the fuel relative to a stationary frae of reference is v v e. Thus, if we equate the total initial oentu of the syste to the total final oentu, we obtain (M )v M(v v) (v v e ) where M represents the ass of the rocket and its reaining fuel after an aount of fuel having ass has been ejected. Siplifying this epression gives We also could have arrived at this result by considering the syste in the center-of-ass frae of reference, which is a frae having the sae velocity as the center of ass of the syste. In this frae, the total oentu of the syste is zero; therefore, if the rocket gains a oentu M v by ejecting soe fuel, the ehausted fuel obtains a oentu v e in the opposite direction, so that M v v e 0. If we now take the liit as t goes to zero, we get v : dv and : d. Futherore, the increase in the ehaust ass d corresponds to an equal decrease in the rocket ass, so that d dm. Note that dm is given a negative sign because it represents a decrease in ass. Using this fact, we obtain M dv v e d v e dm (9.40) Integrating this equation and taking the initial ass of the rocket plus fuel to be M i and the final ass of the rocket plus its reaining fuel to be M f, we obtain vf vi M v v e dv v e Mf Mi dm M v f v i v e ln M i M f (9.4) 5 It is interesting to note that the rocket and achine gun represent cases of the reverse of a perfectly inelastic collision: Moentu is conserved, but the kinetic energy of the syste increases (at the epense of cheical potential energy in the fuel).

17 9.8 Rocket Propulsion CHAPTER 9 Linear Moentu and Collisions This is the basic epression of rocket propulsion. First, it tells us that the increase in rocket speed is proportional to the ehaust speed of the ejected gases, v e. Therefore, the ehaust speed should be very high. Second, the increase in rocket speed is proportional to the natural logarith of the ratio M i /M f. Therefore, this ratio should be as large as possible, which eans that the ass of the rocket without its fuel should be as sall as possible and the rocket should carry as uch fuel as possible. The thrust on the rocket is the force eerted on it by the ejected ehaust gases. We can obtain an epression for the thrust fro Equation 9.40: Thrust M dv (9.42) dt v dm e dt This epression shows us that the thrust increases as the ehaust speed increases and as the rate of change of ass (called the burn rate) increases. EXAMPLE 9.8 A Rocket in Space A rocket oving in free space has a speed of /s relative to the Earth. Its engines are turned on, and fuel is ejected in a direction opposite the rocket s otion at a speed of /s relative to the rocket. (a) What is the speed of the rocket relative to the Earth once the rocket s ass is reduced to one-half its ass before ignition? We can guess that the speed we are looking for ust be greater than the original speed because the rocket is accelerating. Applying Equation 9.4, we obtain v f v i v e ln M i M f EXAMPLE 9.9 Fighting a Fire Two firefighters ust apply a total force of 600 N to steady a hose that is discharging water at L/in. Estiate the speed of the water as it eits the nozzle. The water is eiting at L/in, which is 60 L/s. Knowing that L of water has a ass of kg, we can say that about 60 kg of water leaves the nozzle every second. As the water leaves the hose, it eerts on the hose a thrust that ust be counteracted by the 600-N force eerted on the hose by the firefighters. So, applying Equation 9.42 gives Thrust v e dm dt 600 N v e (60 kg/s) v e 0 /s Firefighting is dangerous work. If the nozzle should slip fro (b) What is the thrust on the rocket if it burns fuel at the rate of 50 kg/s? /s ( /s)ln M i 0.5 M i /s Thrust v e dm dt ( /s)(50 kg/s) N their hands, the oveent of the hose due to the thrust it receives fro the rapidly eiting water could injure the firefighters. Firefighters attack a burning house with a hose line. SUMMARY The linear oentu p of a particle of ass oving with a velocity v is p v (9.) The law of conservation of linear oentu indicates that the total oentu of an isolated syste is conserved. If two particles for an isolated syste, their total oentu is conserved regardless of the nature of the force between the. Therefore, the total oentu of the syste at all ties equals its initial total oentu, or p i p 2i p f p 2f (9.5) The ipulse iparted to a particle by a force F is equal to the change in the oentu of the particle: I tf ti F dt p (9.9) This is known as the ipulse oentu theore. Ipulsive forces are often very strong copared with other forces on the syste and usually act for a very short tie, as in the case of collisions. When two particles collide, the total oentu of the syste before the collision always equals the total oentu after the collision, regardless of the nature of the collision. An inelastic collision is one for which the total kinetic energy is not conserved. A perfectly inelastic collision is one in which the colliding bodies stick together after the collision. An elastic collision is one in which kinetic energy is constant. In a two- or three-diensional collision, the coponents of oentu in each of the three directions (, y, and z) are conserved independently. The position vector of the center of ass of a syste of particles is defined as i r i r CM i (9.30) M where M i is the total ass of the syste and r i is the position vector of the i ith particle. The position vector of the center of ass of a rigid body can be obtained fro the integral epression r CM M r d (9.33) The velocity of the center of ass for a syste of particles is i v i v CM i (9.34) M The total oentu of a syste of particles equals the total ass ultiplied by the velocity of the center of ass. Newton s second law applied to a syste of particles is F et Ma CM dp tot (9.38) dt where a CM is the acceleration of the center of ass and the su is over all eternal forces. The center of ass oves like an iaginary particle of ass M under the

18 Questions CHAPTER 9 Linear Moentu and Collisions influence of the resultant eternal force on the syste. It follows fro Equation 9.38 that the total oentu of the syste is conserved if there are no eternal forces acting on it. PRBLEMS, 2, 3 = straightforward, interediate, challenging = full solution available in the Student s Manual and Study Guide WEB = solution posted at = Coputer useful in solving proble = Interactive Physics = paired nuerical/sybolic probles QUESTINS. If the kinetic energy of a particle is zero, what is its linear oentu? 2. If the speed of a particle is doubled, by what factor is its oentu changed? By what factor is its kinetic energy changed? 3. If two particles have equal kinetic energies, are their oenta necessarily equal? Eplain. 4. If two particles have equal oenta, are their kinetic energies necessarily equal? Eplain. 5. An isolated syste is initially at rest. Is it possible for parts of the syste to be in otion at soe later tie? If so, eplain how this ight occur. 6. If two objects collide and one is initially at rest, is it possible for both to be at rest after the collision? Is it possible for one to be at rest after the collision? Eplain. 7. Eplain how linear oentu is conserved when a ball bounces fro a floor. 8. Is it possible to have a collision in which all of the kinetic energy is lost? If so, cite an eaple. 9. In a perfectly elastic collision between two particles, does the kinetic energy of each particle change as a result of the collision? 0. When a ball rolls down an incline, its linear oentu increases. Does this iply that oentu is not conserved? Eplain.. Consider a perfectly inelastic collision between a car and a large truck. Which vehicle loses ore kinetic energy as a result of the collision? 2. Can the center of ass of a body lie outside the body? If so, give eaples. 3. Three balls are thrown into the air siultaneously. What is the acceleration of their center of ass while they are in otion? 4. A eter stick is balanced in a horizontal position with the inde fingers of the right and left hands. If the two fingers are slowly brought together, the stick reains balanced and the two fingers always eet at the 50-c ark regardless of their original positions (try it!). Eplain. 5. A sharpshooter fires a rifle while standing with the butt of the gun against his shoulder. If the forward oentu of a bullet is the sae as the backward oentu of the gun, why is it not as dangerous to be hit by the gun as by the bullet? 6. A piece of ud is thrown against a brick wall and sticks to the wall. What happens to the oentu of the ud? Is oentu conserved? Eplain. 7. Early in this century, Robert Goddard proposed sending a rocket to the Moon. Critics took the position that in a vacuu, such as eists between the Earth and the Moon, the gases eitted by the rocket would have nothing to push against to propel the rocket. According to Scientific Aerican ( January 975), Goddard placed a gun in a vacuu and fired a blank cartridge fro it. (A blank cartridge fires only the wadding and hot gases of the burning gunpowder.) What happened when the gun was fired? 8. A pole-vaulter falls fro a height of 6.0 onto a foa rubber pad. Can you calculate his speed just before he reaches the pad? Can you estiate the force eerted on hi due to the collision? Eplain. 9. Eplain how you would use a balloon to deonstrate the echanis responsible for rocket propulsion. 20. Does the center of ass of a rocket in free space accelerate? Eplain. Can the speed of a rocket eceed the ehaust speed of the fuel? Eplain. 2. A ball is dropped fro a tall building. Identify the syste for which linear oentu is conserved. 22. A bob, initially at rest, eplodes into several pieces. (a) Is linear oentu conserved? (b) Is kinetic energy conserved? Eplain. 23. NASA often uses the gravity of a planet to slingshot a probe on its way to a ore distant planet. This is actually a collision where the two objects do not touch. How can the probe have its speed increased in this anner? 24. The Moon revolves around the Earth. Is the Moon s linear oentu conserved? Is its kinetic energy conserved? Assue that the Moon s orbit is circular. 25. A raw egg dropped to the floor breaks apart upon ipact. However, a raw egg dropped onto a thick foa rubber cushion fro a height of about rebounds without breaking. Why is this possible? (If you try this eperient, be sure to catch the egg after the first bounce.) 26. n the subject of the following positions, state your own view and argue to support it: (a) The best theory of otion is that force causes acceleration. (b) The true easure of a force s effectiveness is the work it does, and the best theory of otion is that work on an object changes its energy. (c) The true easure of a force s effect is ipulse, and the best theory of otion is that ipulse injected into an object changes its oentu. Section 9. Linear Moentu and Its Conservation. A 3.00-kg particle has a velocity of (3.00i 4.00j) /s. (a) Find its and y coponents of oentu. (b) Find the agnitude and direction of its oentu. 2. A 0.00-kg ball is thrown straight up into the air with an initial speed of 5.0 /s. Find the oentu of the ball (a) at its aiu height and (b) halfway up to its aiu height. 3. A 40.0-kg child standing on a frozen pond throws a kg stone to the east with a speed of 5.00 /s. Neglecting friction between child and ice, find the recoil velocity of the child. 4. A pitcher clais he can throw a baseball with as uch oentu as a 3.00-g bullet oving with a speed of 500 /s. A baseball has a ass of 0.45 kg. What ust be its speed if the pitcher s clai is valid? 5. How fast can you set the Earth oving? In particular, when you jup straight up as high as you can, you give the Earth a aiu recoil speed of what order of agnitude? Model the Earth as a perfectly solid object. In your solution, state the physical quantities you take as data and the values you easure or estiate for the. 6. Two blocks of asses M and 3M are placed on a horizontal, frictionless surface. A light spring is attached to one of the, and the blocks are pushed together with the spring between the (Fig. P9.6). A cord initially holding the blocks together is burned; after this, the block of ass 3M oves to the right with a speed of 2.00 /s. (a) What is the speed of the block of ass M? (b) Find the original elastic energy in the spring if M kg. M v M Before (a) After (b) Figure P9.6 3M 2.00 /s 3M WEB 7. (a) A particle of ass oves with oentu p. Show that the kinetic energy of the particle is given by K p 2 /2. (b) Epress the agnitude of the particle s oentu in ters of its kinetic energy and ass. Section 9.2 Ipulse and Moentu 8. A car is stopped for a traffic signal. When the light turns green, the car accelerates, increasing its speed fro zero to 5.20 /s in s. What linear ipulse and average force does a 70.0-kg passenger in the car eperience? 9. An estiated force tie curve for a baseball struck by a bat is shown in Figure P9.9. Fro this curve, deterine (a) the ipulse delivered to the ball, (b) the average force eerted on the ball, and (c) the peak force eerted on the ball. F(N) F = N t(s) Figure P A tennis player receives a shot with the ball ( kg) traveling horizontally at 50.0 /s and returns the shot with the ball traveling horizontally at 40.0 /s in the opposite direction. (a) What is the ipulse delivered to the ball by the racket? (b) What work does the racket do on the ball?. A 3.00-kg steel ball strikes a wall with a speed of 0.0 /s at an angle of 60.0 with the surface. It bounces off with the sae speed and angle (Fig. P9.). If the ball is in contact with the wall for s, what is the average force eerted on the ball by the wall? 2. In a slow-pitch softball gae, a kg softball crossed the plate at 5.0 /s at an angle of 45.0 below the horizontal. The ball was hit at 40.0 /s, 30.0 above the horizontal. (a) Deterine the ipulse delivered to the ball. (b) If the force on the ball increased linearly for 4.00 s, held constant for 20.0 s, and then decreased to zero linearly in another 4.00 s, what was the aiu force on the ball?

19 Probles CHAPTER 9 Linear Moentu and Collisions Figure P9. 3. A garden hose is held in the anner shown in Figure P9.3. The hose is initially full of otionless water. What additional force is necessary to hold the nozzle stationary after the water is turned on if the discharge rate is kg/s with a speed of 25.0 /s? Figure P A professional diver perfors a dive fro a platfor 0 above the water surface. Estiate the order of agnitude of the average ipact force she eperiences in her collision with the water. State the quantities you take as data and their values. Section 9.3 Collisions Section 9.4 Elastic and Inelastic Collisions in ne Diension 5. High-speed stroboscopic photographs show that the head of a golf club of ass 200 g is traveling at 55.0 /s just before it strikes a 46.0-g golf ball at rest on a tee. After the collision, the club head travels (in the sae direction) at 40.0 /s. Find the speed of the golf ball just after ipact. 6. A 75.0-kg ice skater, oving at 0.0 /s, crashes into a stationary skater of equal ass. After the collision, the two skaters ove as a unit at 5.00 /s. Suppose the average force a skater can eperience without breaking a bone is N. If the ipact tie is 0.00 s, does a bone break? 7. A 0.0-g bullet is fired into a stationary block of wood ( 5.00 kg). The relative otion of the bullet stops y inside the block. The speed of the bullet-plus-wood cobination iediately after the collision is easured as /s. What was the original speed of the bullet? 8. As shown in Figure P9.8, a bullet of ass and speed v passes copletely through a pendulu bob of ass M. The bullet eerges with a speed of v/2. The pendulu bob is suspended by a stiff rod of length and negligible ass. What is the iniu value of v such that the pendulu bob will barely swing through a coplete vertical circle? M v v/2 Figure P A 45.0-kg girl is standing on a plank that has a ass of 50 kg. The plank, originally at rest, is free to slide on a frozen lake, which is a flat, frictionless supporting surface. The girl begins to walk along the plank at a constant speed of.50 /s relative to the plank. (a) What is her speed relative to the ice surface? (b) What is the speed of the plank relative to the ice surface? 20. Gayle runs at a speed of 4.00 /s and dives on a sled, which is initially at rest on the top of a frictionless snowcovered hill. After she has descended a vertical distance of 5.00, her brother, who is initially at rest, hops on her back and together they continue down the hill. What is their speed at the botto of the hill if the total vertical drop is 5.0? Gayle s ass is 50.0 kg, the sled has a ass of 5.00 kg and her brother has a ass of 30.0 kg. 2. A 200-kg car traveling initially with a speed of 25.0 /s in an easterly direction crashes into the rear end of a kg truck oving in the sae direction at 20.0 /s (Fig. P9.2). The velocity of the car right after the collision is 8.0 /s to the east. (a) What is the velocity of the truck right after the collision? (b) How uch echanical energy is lost in the collision? Account for this loss in energy. 22. A railroad car of ass kg is oving with a speed of 4.00 /s. It collides and couples with three other coupled railroad cars, each of the sae ass as the single car and oving in the sae direction with an initial speed of 2.00 /s. (a) What is the speed of the four cars after the collision? (b) How uch energy is lost in the collision? WEB WEB 25.0 /s 20.0 /s 8.0 /s 23. Four railroad cars, each of ass kg, are coupled together and coasting along horizontal tracks at a speed of v i toward the south. A very strong but foolish ovie actor, riding on the second car, uncouples the front car and gives it a big push, increasing its speed to 4.00 /s southward. The reaining three cars continue oving toward the south, now at 2.00 /s. (a) Find the initial speed of the cars. (b) How uch work did the actor do? (c) State the relationship between the process described here and the process in Proble A 7.00-kg bowling ball collides head-on with a 2.00-kg bowling pin. The pin flies forward with a speed of 3.00 /s. If the ball continues forward with a speed of.80 /s, what was the initial speed of the ball? Ignore rotation of the ball. 25. A neutron in a reactor akes an elastic head-on collision with the nucleus of a carbon ato initially at rest. (a) What fraction of the neutron s kinetic energy is transferred to the carbon nucleus? (b) If the initial kinetic energy of the neutron is J, find its final kinetic energy and the kinetic energy of the carbon nucleus after the collision. (The ass of the carbon nucleus is about 2.0 ties greater than the ass of the neutron.) 26. Consider a frictionless track ABC as shown in Figure P9.26. A block of ass 5.00 kg is released fro A. It akes a head-on elastic collision at B with a block of ass kg that is initially at rest. Calculate the aiu height to which rises after the collision. A 5.00 B 2 Figure P9.26 Before BIG Joes IRISH BEER 27. A 2.0-g bullet is fired into a 00-g wooden block initially at rest on a horizontal surface. After ipact, the block slides 7.50 before coing to rest. If the coefficient of friction between the block and the surface is Figure P9.2 C v BIG Joes IRISH BEER After 0.650, what was the speed of the bullet iediately before ipact? 28. A 7.00-g bullet, when fired fro a gun into a.00-kg block of wood held in a vise, would penetrate the block to a depth of 8.00 c. This block of wood is placed on a frictionless horizontal surface, and a 7.00-g bullet is fired fro the gun into the block. To what depth will the bullet penetrate the block in this case? Section 9.5 Two-Diensional Collisions 29. A 90.0-kg fullback running east with a speed of 5.00 /s is tackled by a 95.0-kg opponent running north with a speed of 3.00 /s. If the collision is perfectly inelastic, (a) calculate the speed and direction of the players just after the tackle and (b) deterine the energy lost as a result of the collision. Account for the issing energy. 30. The ass of the blue puck in Figure P9.30 is 20.0% greater than the ass of the green one. Before colliding, the pucks approach each other with equal and opposite oenta, and the green puck has an initial speed of 0.0 /s. Find the speeds of the pucks after the collision if half the kinetic energy is lost during the collision Figure P Two autoobiles of equal ass approach an intersection. ne vehicle is traveling with velocity 3.0 /s toward the east and the other is traveling north with a speed of v 2i. Neither driver sees the other. The vehicles collide in the intersection and stick together, leaving parallel skid arks at an angle of 55.0 north of east. The speed liit for both roads is 35 i/h, and the driver of the northward-oving vehicle clais he was within the speed liit when the collision occurred. Is he telling the truth?

20 Probles CHAPTER 9 Linear Moentu and Collisions WEB 32. A proton, oving with a velocity of v i i, collides elastically with another proton that is initially at rest. If the two protons have equal speeds after the collision, find (a) the speed of each proton after the collision in ters of v i and (b) the direction of the velocity vectors after the collision. 33. A billiard ball oving at 5.00 /s strikes a stationary ball of the sae ass. After the collision, the first ball oves at 4.33 /s and at an angle of 30.0 with respect to the original line of otion. Assuing an elastic collision (and ignoring friction and rotational otion), find the struck ball s velocity. 34. A kg puck, initially at rest on a horizontal, frictionless surface, is struck by a kg puck oving initially along the ais with a speed of 2.00 /s. After the collision, the kg puck has a speed of.00 /s at an angle of 53.0 to the positive ais (see Fig. 9.4). (a) Deterine the velocity of the kg puck after the collision. (b) Find the fraction of kinetic energy lost in the collision. 35. A 3.00-kg ass with an initial velocity of 5.00i /s collides with and sticks to a 2.00-kg ass with an initial velocity of 3.00j /s. Find the final velocity of the coposite ass. 36. Two shuffleboard disks of equal ass, one orange and the other yellow, are involved in an elastic, glancing collision. The yellow disk is initially at rest and is struck by the orange disk oving with a speed of 5.00 /s. After the collision, the orange disk oves along a direction that akes an angle of 37.0 with its initial direction of otion, and the velocity of the yellow disk is perpendicular to that of the orange disk (after the collision). Deterine the final speed of each disk. 37. Two shuffleboard disks of equal ass, one orange and the other yellow, are involved in an elastic, glancing collision. The yellow disk is initially at rest and is struck by the orange disk oving with a speed v i. After the collision, the orange disk oves along a direction that akes an angle with its initial direction of otion, and the velocity of the yellow disk is perpendicular to that of the orange disk (after the collision). Deterine the final speed of each disk. 38. During the battle of Gettysburg, the gunfire was so intense that several bullets collided in idair and fused together. Assue a 5.00-g Union usket ball was oving to the right at a speed of 250 /s, 20.0 above the horizontal, and that a 3.00-g Confederate ball was oving to the left at a speed of 280 /s, 5.0 above the horizontal. Iediately after they fuse together, what is their velocity? 39. An unstable nucleus of ass kg initially at rest disintegrates into three particles. ne of the particles, of ass kg, oves along the y ais with a velocity of /s. Another particle, of ass kg, oves along the ais with a speed of /s. Find (a) the velocity of the third particle and (b) the total kinetic energy increase in the process. Section 9.6 The Center of Mass 40. Four objects are situated along the y ais as follows: A 2.00-kg object is at 3.00, a 3.00-kg object is at 2.50, a 2.50-kg object is at the origin, and a 4.00-kg object is at Where is the center of ass of these objects? 4. A unifor piece of sheet steel is shaped as shown in Figure P9.4. Copute the and y coordinates of the center of ass of the piece. y(c) Figure P The ass of the Earth is kg, and the ass of the Moon is kg. The distance of separation, easured between their centers, is Locate the center of ass of the Earth Moon syste as easured fro the center of the Earth. 43. A water olecule consists of an oygen ato with two hydrogen atos bound to it (Fig. P9.43). The angle between the two bonds is 06. If the bonds are 0.00 n long, where is the center of ass of the olecule? 0.00 n 0.00 n Figure P9.43 H H (c) 44. A kg ass has position r 2.0j c. A kg ass 2 has position r 2 2.0i c. Another kg ass 3 has position r 3 (2.0i 2.0j) c. Make a drawing of the asses. Start fro the origin and, to the scale c kg c, construct the vector r, then the vector r 2 r 2, then the vector r 2 r 2 3 r 3, and at last r CM ( r 2 r 2 3 r 3 )/( 2 3 ). bserve that the head of the vector r CM indicates the position of the center of ass. 45. A rod of length 30.0 c has linear density (ass-perlength) given by where is the distance fro one end, easured in eters. (a) What is the ass of the rod? (b) How far fro the 0 end is its center of ass? Section g/ 20.0 g/ 2 Motion of a Syste of Particles 46. Consider a syste of two particles in the y plane: 2.00 kg is at r (.00i 2.00j) and has velocity (3.00i 0.500j) /s; kg is at r 2 ( 4.00i 3.00j) and has velocity (3.00i 2.00j) /s. (a) Plot these particles on a grid or graph paper. Draw their position vectors and show their velocities. (b) Find the position of the center of ass of the syste and ark it on the grid. (c) Deterine the velocity of the center of ass and also show it on the diagra. (d) What is the total linear oentu of the syste? 47. Roeo (77.0 kg) entertains Juliet (55.0 kg) by playing his guitar fro the rear of their boat at rest in still water, 2.70 away fro Juliet who is in the front of the boat. After the serenade, Juliet carefully oves to the rear of the boat (away fro shore) to plant a kiss on Roeo s cheek. How far does the 80.0-kg boat ove toward the shore it is facing? 48. Two asses, kg and kg, begin unifor otion at the sae speed, /s, fro the origin at t 0 and travel in the directions shown in Figure P9.48. (a) Find the velocity of the center of ass in unit vector notation. (b) Find the agnitude and direction y kg kg Figure P9.48 of the velocity of the center of ass. (c) Write the position vector of the center of ass as a function of tie. 49. A 2.00-kg particle has a velocity of (2.00i 3.00j) /s, and a 3.00-kg particle has a velocity of (.00i 6.00j) /s. Find (a) the velocity of the center of ass and (b) the total oentu of the syste. 50. A ball of ass kg has a velocity of.50i /s; a ball of ass kg has a velocity of 0.400i /s. They eet in a head-on elastic collision. (a) Find their velocities after the collision. (b) Find the velocity of their center of ass before and after the collision. (ptional) Section 9.8 Rocket Propulsion 5. The first stage of a Saturn V space vehicle consues fuel and oidizer at the rate of kg/s, with an ehaust speed of /s. (a) Calculate the thrust produced by these engines. (b) Find the initial acceleration of the vehicle on the launch pad if its initial ass is kg. [Hint: You ust include the force of gravity to solve part (b).] 52. A large rocket with an ehaust speed of v e /s develops a thrust of 24.0 illion newtons. (a) How uch ass is being blasted out of the rocket ehaust per second? (b) What is the aiu speed the rocket can attain if it starts fro rest in a force-free environent with v e 3.00 k/s and if 90.0% of its initial ass is fuel and oidizer? 53. A rocket for use in deep space is to have the capability of boosting a total load (payload plus rocket frae and engine) of 3.00 etric tons to a speed of /s. (a) It has an engine and fuel designed to produce an ehaust speed of /s. How uch fuel plus oidizer is required? (b) If a different fuel and engine design could give an ehaust speed of /s, what aount of fuel and oidizer would be required for the sae task? 54. A rocket car has a ass of kg unfueled and a ass of kg when copletely fueled. The ehaust velocity is /s. (a) Calculate the aount of fuel used to accelerate the copletely fueled car fro rest to 225 /s (about 500 i/h). (b) If the burn rate is constant at 30.0 kg/s, calculate the tie it takes the car to reach this speed. Neglect friction and air resistance. WEB ADDITINAL PRBLEMS 55. Review Proble. A 60.0-kg person running at an initial speed of 4.00 /s jups onto a 20-kg cart initially at rest (Fig. P9.55). The person slides on the cart s top surface and finally coes to rest relative to the cart. The coefficient of kinetic friction between the person and the cart is Friction between the cart and ground can be neglected. (a) Find the final velocity of the person and cart relative to the ground. (b) Find the frictional force acting on the person while he is sliding

21 Probles CHAPTER 9 Linear Moentu and Collisions across the top surface of the cart. (c) How long does the frictional force act on the person? (d) Find the change in oentu of the person and the change in oentu of the cart. (e) Deterine the displaceent of the person relative to the ground while he is sliding on the cart. (f) Deterine the displaceent of the cart relative to the ground while the person is sliding. (g) Find the change in kinetic energy of the person. (h) Find the change in kinetic energy of the cart. (i) Eplain why the answers to parts (g) and (h) differ. (What kind of collision is this, and what accounts for the loss of echanical energy?) 60.0 kg 4.00 /s Figure P kg 56. A golf ball ( 46.0 g) is struck a blow that akes an angle of 45.0 with the horizontal. The ball lands 200 away on a flat fairway. If the golf club and ball are in contact for 7.00 s, what is the average force of ipact? (Neglect air resistance.) 57. An 8.00-g bullet is fired into a 2.50-kg block that is initially at rest at the edge of a frictionless table of height.00 (Fig. P9.57). The bullet reains in the block, and after ipact the block lands 2.00 fro the botto of the table. Deterine the initial speed of the bullet. 58. A bullet of ass is fired into a block of ass M that is initially at rest at the edge of a frictionless table of height h (see Fig. P9.57). The bullet reains in the block, and after ipact the block lands a distance d fro the botto of the table. Deterine the initial speed of the bullet. 59. An 80.0-kg astronaut is working on the engines of his ship, which is drifting through space with a constant velocity. The astronaut, wishing to get a better view of the Universe, pushes against the ship and uch later finds hiself 30.0 behind the ship and at rest with respect to it. Without a thruster, the only way to return to the ship is to throw his kg wrench directly away fro the ship. If he throws the wrench with a speed of 20.0 /s relative to the ship, how long does it take the astronaut to reach the ship? 60. A sall block of ass kg is released fro rest at the top of a curve-shaped frictionless wedge of ass kg, which sits on a frictionless horizontal surface, as shown in Figure P9.60a. When the block leaves the wedge, its velocity is easured to be 4.00 /s to the right, as in Figure P9.60b. (a) What is the velocity of the wedge after the block reaches the horizontal surface? (b) What is the height h of the wedge? h 2 (a) v 2 Figure P (b) 4.00 /s 6. Tarzan, whose ass is 80.0 kg, swings fro a vine that is horizontal when he starts. At the botto of his arc, he picks up 60.0-kg Jane in a perfectly inelastic collision. What is the height of the highest tree lib they can reach on their upward swing? 62. A jet aircraft is traveling at 500 i/h (223 /s) in horizontal flight. The engine takes in air at a rate of 80.0 kg/s and burns fuel at a rate of 3.00 kg/s. If the ehaust gases are ejected at 600 /s relative to the aircraft, find the thrust of the jet engine and the delivered horsepower. 63. A 75.0-kg firefighter slides down a pole while a constant frictional force of 300 N retards her otion. A horizontal 20.0-kg platfor is supported by a spring at the botto of the pole to cushion the fall. The firefighter starts fro rest 4.00 above the platfor, and the spring constant is N/. Find (a) the firefighter s speed just before she collides with the platfor and (b) the aiu distance the spring is copressed. (Assue the frictional force acts during the entire otion.) 64. A cannon is rigidly attached to a carriage, which can ove along horizontal rails but is connected to a post by a large spring, initially unstretched and with force constant k N/, as shown in Figure P9.64. The cannon fires a 200-kg projectile at a velocity of 25 /s directed 45.0 above the horizontal. (a) If the ass of the cannon and its carriage is kg, find the recoil speed of the cannon. (b) Deterine the aiu etension of the spring. (c) Find the aiu force the spring eerts on the carriage. (d) Consider the syste consisting of the cannon, carriage, and shell. Is the oentu of this syste conserved during the firing? Why or why not? (a) 66. Two gliders are set in otion on an air track. A spring of force constant k is attached to the near side of one glider. The first glider of ass has a velocity of v, and the second glider of ass 2 has a velocity of v 2, as shown in Figure P9.66 (v v 2 ). When collides with the spring attached to 2 and copresses the spring to its aiu copression, the velocity of the gliders is v. In ters of v, v 2,, 2, and k, find (a) the velocity v at aiu copression, (b) the aiu copression, and (c) the velocities of each glider after has lost contact with the spring. v L Figure P9.65 k v 2 (b) 2 L 8.00 g 2.50 kg 45.0 Figure P Figure P9.57 Probles 57 and 58. Figure P A chain of length L and total ass M is released fro rest with its lower end just touching the top of a table, as shown in Figure P9.65a. Find the force eerted by the table on the chain after the chain has fallen through a distance, as shown in Figure P9.65b. (Assue each link coes to rest the instant it reaches the table.) 67. Sand fro a stationary hopper falls onto a oving conveyor belt at the rate of 5.00 kg/s, as shown in Figure P9.67. The conveyor belt is supported by frictionless rollers and oves at a constant speed of /s under the action of a constant horizontal eternal force F et supplied by the otor that drives the belt. Find (a) the sand s rate of change of oentu in the horizontal direction, (b) the force of friction eerted by the belt on the sand, (c) the eternal force F et, (d) the work done by F et in s, and (e) the kinetic energy acquired by the falling sand each second due to the change in its horizontal otion. (f) Why are the answers to parts (d) and (e) different?

22 Probles CHAPTER 9 Linear Moentu and Collisions 68. A rocket has total ass M i 360 kg, including 330 kg of fuel and oidizer. In interstellar space it starts fro rest, turns on its engine at tie t 0, and puts out ehaust with a relative speed of v e 500 /s at the constant rate k 2.50 kg/s. Although the fuel will last for an actual burn tie of 330 kg/(2.5 kg/s) 32 s, define a projected depletion tie as T p M i /k 360 kg/(2.5 kg/s) 44 s. (This would be the burn tie if the rocket could use its payload, fuel tanks, and even the walls of the cobustion chaber as fuel.) (a) Show that during the burn the velocity of the rocket is given as a function of tie by v(t) v e ln( t/t p ) (b) Make a graph of the velocity of the rocket as a function of tie for ties running fro 0 to 32 s. (c) Show that the acceleration of the rocket is a(t) v e /(T p t) (d) Graph the acceleration as a function of tie. (e) Show that the displaceent of the rocket fro its initial position at t 0 is (f) Graph the displaceent during the burn. 69. A 40.0-kg child stands at one end of a 70.0-kg boat that is 4.00 in length (Fig. P9.69). The boat is initially 3.00 fro the pier. The child notices a turtle on a rock near the far end of the boat and proceeds to walk to that end to catch the turtle. Neglecting friction be(t) v e (T p t)ln( t/t p ) v e t Figure P /s Figure P9.67 F et tween the boat and the water, (a) describe the subsequent otion of the syste (child plus boat). (b) Where is the child relative to the pier when he reaches the far end of the boat? (c) Will he catch the turtle? (Assue he can reach out.00 fro the end of the boat.) 70. A student perfors a ballistic pendulu eperient, using an apparatus siilar to that shown in Figure 9.b. She obtains the following average data: h 8.68 c, 68.8 g, and g. The sybols refer to the quantities in Figure 9.a. (a) Deterine the initial speed v i of the projectile. (b) In the second part of her eperient she is to obtain v i by firing the sae projectile horizontally (with the pendulu reoved fro the path) and easuring its horizontal displaceent and vertical displaceent y (Fig. P9.70). Show that the initial speed of the projectile is related to and y through the relationship v i 2y/g What nuerical value does she obtain for v i on the basis of her easured values of 257 c and y 85.3 c? What factors ight account for the difference in this value copared with that obtained in part (a)? y v i Figure P A 5.00-g bullet oving with an initial speed of 400 /s is fired into and passes through a.00-kg block, as shown in Figure P9.7. The block, initially at rest on a frictionless, horizontal surface, is connected to a spring of force constant 900 N/. If the block oves 5.00 c to the right after ipact, find (a) the speed at which the bullet eerges fro the block and (b) the energy lost in the collision. 400 /s 5.00 c Figure P Two asses and 3 are oving toward each other along the ais with the sae initial speeds v i. Mass is traveling to the left, while ass 3 is traveling to the right. They undergo a head-on elastic collision and each rebounds along the sae line as it approached. Find the final speeds of the asses. 73. Two asses and 3 are oving toward each other along the ais with the sae initial speeds v i. Mass is traveling to the left, while ass 3 is traveling to the right. They undergo an elastic glancing collision such ANSWERS T QUICK QUIZZES 9. (d). Two identical objects ( 2 ) traveling in the sae direction at the sae speed (v v 2 ) have the sae kinetic energies and the sae oenta. However, this is not true if the two objects are oving at the sae speed but in different directions. In the latter case, K K 2, but the differing velocity directions indicate that p p 2 because oentu is a vector quantity. It also is possible for particular cobinations of asses and velocities to satisfy K K 2 but not p p 2. For eaple, a -kg object oving at 2 /s has the sae kinetic energy as a 4-kg object oving at /s, but the two clearly do not have the sae oenta. 9.2 (b), (c), (a). The slower the ball, the easier it is to catch. If the oentu of the edicine ball is the sae as the oentu of the baseball, the speed of the edicine ball ust be /0 the speed of the baseball because the edicine ball has 0 ties the ass. If the kinetic energies are the sae, the speed of the edicine ball ust be / 0 the speed of the baseball because of the squared speed ter in the forula for K. The edicine v that ass is oving downward after the collision at right angles fro its initial direction. (a) Find the final speeds of the two asses. (b) What is the angle at which the ass 3 is scattered? 74. Review Proble. There are (one can say) three coequal theories of otion: Newton s second law, stating that the total force on an object causes its acceleration; the work kinetic energy theore, stating that the total work on an object causes its change in kinetic energy; and the ipulse oentu theore, stating that the total ipulse on an object causes its change in oentu. In this proble, you copare predictions of the three theories in one particular case. A 3.00-kg object has a velocity of 7.00j /s. Then, a total force 2.0i N acts on the object for 5.00 s. (a) Calculate the object s final velocity, using the ipulse oentu theore. (b) Calculate its acceleration fro a (v f v i )/t. (c) Calculate its acceleration fro a F/. (d) Find the object s vector displaceent fro r v i t 2 a t 2. (e) Find the work done on the object fro W F r. (f) Find the final kinetic energy fro 2 v f 2 2 v f v f. (g) Find the final kinetic energy fro 2 v i 2 W. 75. A rocket has a total ass of M i 360 kg, including 330 kg of fuel and oidizer. In interstellar space it starts fro rest. Its engine is turned on at tie t 0, and it puts out ehaust with a relative speed of v e 500 /s at the constant rate 2.50 kg/s. The burn lasts until the fuel runs out at tie 330 kg/(2.5 kg/s) 32 s. Set up and carry out a coputer analysis of the otion according to Euler s ethod. Find (a) the final velocity of the rocket and (b) the distance it travels during the burn. ball is hardest to catch when it has the sae speed as the baseball. 9.3 (c) and (e). bject 2 has a greater acceleration because of its saller ass. Therefore, it takes less tie to travel the distance d. Thus, even though the force applied to objects and 2 is the sae, the change in oentu is less for object 2 because t is saller. Therefore, because the initial oenta were the sae (both zero), p p 2. The work W Fd done on both objects is the sae because both F and d are the sae in the two cases. Therefore, K K Because the passenger is brought fro the car s initial speed to a full stop, the change in oentu (the ipulse) is the sae regardless of whether the passenger is stopped by dashboard, seatbelt, or airbag. However, the dashboard stops the passenger very quickly in a frontend collision. The seatbelt takes soewhat ore tie. Used along with the seatbelt, the airbag can etend the passenger s stopping tie further, notably for his head, which would otherwise snap forward. Therefore, the

23 Probles 29 dashboard applies the greatest force, the seatbelt an interediate force, and the airbag the least force. Airbags are designed to work in conjunction with seatbelts. Make sure you wear your seatbelt at all ties while in a oving vehicle. 9.5 If we define the ball as our syste, oentu is not conserved. The ball s speed and hence its oentu continually increase. This is consistent with the fact that the gravitational force is eternal to this chosen syste. However, if we define our syste as the ball and the Earth, oentu is conserved, for the Earth also has oentu because the ball eerts a gravitational force on it. As the ball falls, the Earth oves up to eet it (although the Earth s speed is on the order of 0 25 ties less than that of the ball!). This upward oveent changes the Earth s oentu. The change in the Earth s oentu is nuerically equal to the change in the ball s oentu but is in the opposite direction. Therefore, the total oentu of the Earth ball syste is conserved. Because the Earth s ass is so great, its upward otion is negligibly sall. 9.6 (c). The greatest ipulse (greatest change in oentu) is iparted to the Frisbee when the skater reverses its oentu vector by catching it and throwing it back. Since this is when the skater iparts the greatest ipulse to the Frisbee, then this also is when the Frisbee iparts the greatest ipulse to her. 9.7 Both are equally bad. Iagine watching the collision fro a safer location alongside the road. As the crush zones of the two cars are copressed, you will see that the actual point of contact is stationary. You would see the sae thing if your car were to collide with a solid wall. 9.8 No, such oveent can never occur if we assue the collisions are elastic. The oentu of the syste before the collision is v, where is the ass of ball and v is its speed just before the collision. After the collision, we would have two balls, each of ass and oving with a speed of v/2. Thus, the total oentu of the syste after the collision would be (v/2) (v/2) v. Thus, oentu is conserved. However, the kinetic energy just before the collision is K i 2 v 2, and that after the collision is K f 2 (v/2)2 2 (v/2)2 4 v 2. Thus, kinetic energy is not conserved. Both oentu and kinetic energy are conserved only when one ball oves out when one ball is released, two balls ove out when two are released, and so on. 9.9 No they will not! The piece with the handle will have less ass than the piece ade up of the end of the bat. To see why this is so, take the origin of coordinates as the center of ass before the bat was cut. Replace each cut piece by a sall sphere located at the center of ass for each piece. The sphere representing the handle piece is farther fro the origin, but the product of lesser ass and greater distance balances the product of greater ass and lesser distance for the end piece:

24 P U Z Z L E R Did you know that the CD inside this player spins at different speeds, depending on which song is playing? Why would such a strange characteristic be incorporated into the design of every CD player? (George Seple) 0. Angular Displaceent, Velocity, and Acceleration 293 When an etended object, such as a wheel, rotates about its ais, the otion cannot be analyzed by treating the object as a particle because at any given tie different parts of the object have different linear velocities and linear accelerations. For this reason, it is convenient to consider an etended object as a large nuber of particles, each of which has its own linear velocity and linear acceleration. In dealing with a rotating object, analysis is greatly siplified by assuing that the object is rigid. A rigid object is one that is nondeforable that is, it is an object in which the separations between all pairs of particles reain constant. All real bodies are deforable to soe etent; however, our rigid-object odel is useful in any situations in which deforation is negligible. In this chapter, we treat the rotation of a rigid object about a fied ais, which is coonly referred to as pure rotational otion. Rigid object 0. ANGULAR DISPLACEMENT, VELCITY, AND ACCELERATIN c h a p t e r Rotation of a Rigid bject About a Fied Ais Figure 0. illustrates a planar (flat), rigid object of arbitrary shape confined to the y plane and rotating about a fied ais through. The ais is perpendicular to the plane of the figure, and is the origin of an y coordinate syste. Let us look at the otion of only one of the illions of particles aking up this object. A particle at P is at a fied distance r fro the origin and rotates about it in a circle of radius r. (In fact, every particle on the object undergoes circular otion about.) It is convenient to represent the position of P with its polar coordinates (r, ), where r is the distance fro the origin to P and is easured counterclockwise fro soe preferred direction in this case, the positive ais. In this representation, the only coordinate that changes in tie is the angle ; r reains constant. (In cartesian coordinates, both and y vary in tie.) As the particle oves along the circle fro the positive ais ( 0) to P, it oves through an arc of length s, which is related to the angular position through the relationship s r s r (0.a) (0.b) It is iportant to note the units of in Equation 0.b. Because is the ratio of an arc length and the radius of the circle, it is a pure nuber. However, we coonly give the artificial unit radian (rad), where y Figure 0. A rigid object rotating about a fied ais through perpendicular to the plane of the figure. (In other words, the ais of rotation is the z ais.) A particle at P rotates in a circle of radius r centered at. r θ P s Chapter utline one radian is the angle subtended by an arc length equal to the radius of the arc. Radian 0. Angular Displaceent, Velocity, and Acceleration 0.2 Rotational Kineatics: Rotational Motion with Constant Angular Acceleration 0.3 Angular and Linear Quantities 0.4 Rotational Energy 0.5 Calculation of Moents of Inertia 0.6 Torque 0.7 Relationship Between Torque and Angular Acceleration 0.8 Work, Power, and Energy in Rotational Motion Because the circuference of a circle is 2r, it follows fro Equation 0.b that 360 corresponds to an angle of 2r/r rad 2 rad (one revolution). Hence, rad 360 / To convert an angle in degrees to an angle in radians, we use the fact that 2 rad 360 : (rad) For eaple, 60 equals /3 rad, and 45 equals /4 rad. 80 (deg) 292

25 294 CHAPTER 0 Rotation of a Rigid bject About a Fied Ais 0. Angular Displaceent, Velocity, and Acceleration 295 ω y Q,t f ω Figure 0.3 The right-hand rule for deterining the direction of the angular velocity vector. r P, t i θ f θ i Figure 0.2 A particle on a rotating rigid object oves fro P to Q along the arc of a circle. In the tie interval t t f t i, the radius vector sweeps out an angle f i. Average angular speed Instantaneous angular speed Average angular acceleration As the particle in question on our rigid object travels fro position P to position Q in a tie t as shown in Figure 0.2, the radius vector sweeps out an angle f i. This quantity is defined as the angular displaceent of the particle: (0.2) We define the average angular speed (oega) as the ratio of this angular displaceent to the tie interval t: f i (0.3) t f t i t In analogy to linear speed, the instantaneous angular speed is defined as the liit of the ratio /t as t approaches zero: (0.4) Angular speed has units of radians per second (rad/s), or rather second (s ) because radians are not diensional. We take to be positive when is increasing (counterclockwise otion) and negative when is decreasing (clockwise otion). If the instantaneous angular speed of an object changes fro i to f in the tie interval t, the object has an angular acceleration. The average angular acceleration (alpha) of a rotating object is defined as the ratio of the change in the angular speed to the tie interval t : f i (0.5) t f t i t li f i t:0 t In a short track event, such as a 200- or 400- sprint, the runners begin fro staggered positions on the track. Why don t they all begin fro the sae line? d dt In analogy to linear acceleration, the instantaneous angular acceleration is defined as the liit of the ratio /t as t approaches zero: d (0.6) t:0 t dt Angular acceleration has units of radians per second squared (rad/s 2 ), or just second 2 (s 2 ). Note that is positive when the rate of counterclockwise rotation is increasing or when the rate of clockwise rotation is decreasing. When rotating about a fied ais, every particle on a rigid object rotates through the sae angle and has the sae angular speed and the sae angular acceleration. That is, the quantities,, and characterize the rotational otion of the entire rigid object. Using these quantities, we can greatly siplify the analysis of rigid-body rotation. Angular position (), angular speed (), and angular acceleration () are analogous to linear position (), linear speed (v), and linear acceleration (a). The variables,, and differ diensionally fro the variables, v, and a only by a factor having the unit of length. We have not specified any direction for and. Strictly speaking, these variables are the agnitudes of the angular velocity and the angular acceleration vectors and, respectively, and they should always be positive. Because we are considering rotation about a fied ais, however, we can indicate the directions of the vectors by assigning a positive or negative sign to and, as discussed earlier with regard to Equations 0.4 and 0.6. For rotation about a fied ais, the only direction that uniquely specifies the rotational otion is the direction along the ais of rotation. Therefore, the directions of and are along this ais. If an object rotates in the y plane as in Figure 0., the direction of is out of the plane of the diagra when the rotation is counterclockwise and into the plane of the diagra when the rotation is clockwise. To illustrate this convention, it is convenient to use the right-hand rule deonstrated in Figure 0.3. When the four fingers of the right hand are wrapped in the direction of rotation, the etended right thub points in the direction of. The direction of follows fro its definition d/dt. It is the sae as the direction of if the angular speed is increasing in tie, and it is antiparallel to if the angular speed is decreasing in tie. li Instantaneous angular acceleration

26 296 CHAPTER 0 Rotation of a Rigid bject About a Fied Ais 0.3 Angular and Linear Quantities 297 Rotational kineatic equations EXAMPLE 0. Quick Quiz 0. Describe a situation in which 0 and and are antiparallel. 0.2 RTATINAL KINEMATICS: RTATINAL MTIN WITH CNSTANT ANGULAR ACCELERATIN In our study of linear otion, we found that the siplest for of accelerated otion to analyze is otion under constant linear acceleration. Likewise, for rotational otion about a fied ais, the siplest accelerated otion to analyze is otion under constant angular acceleration. Therefore, we net develop kineatic relationships for this type of otion. If we write Equation 0.6 in the for d dt, and let t i 0 and t f t, we can integrate this epression directly: f i t (for constant ) (0.7) Substituting Equation 0.7 into Equation 0.4 and integrating once ore we obtain f i it 2t 2 If we eliinate t fro Equations 0.7 and 0.8, we obtain f 2 i 2 2(f i) (for constant ) (0.8) (for constant ) (0.9) Notice that these kineatic epressions for rotational otion under constant angular acceleration are of the sae for as those for linear otion under constant linear acceleration with the substitutions :, v :, and a :. Table 0. copares the kineatic equations for rotational and linear otion. Rotating Wheel A wheel rotates with a constant angular acceleration of 3.50 rad/s 2. If the angular speed of the wheel is 2.00 rad/s at t i 0, (a) through what angle does the wheel rotate in 2.00 s? We can use Figure 0.2 to represent the wheel, and so we do not need a new drawing. This is a straightforward application of an equation fro Table 0.: f i it 2t 2 (2.00 rad/s)(2.00 s) 2 (3.50 rad/s2 )(2.00 s) 2.0 rad (.0 rad)(57.3 /rad) /rev.75 rev (b) What is the angular speed at t 2.00 s? Because the angular acceleration and the angular speed are both positive, we can be sure our answer ust be greater than 2.00 rad/s. We could also obtain this result using Equation 0.9 and the results of part (a). Try it! You also ay want to see if you can forulate the linear otion analog to this proble. Eercise Find the angle through which the wheel rotates between t 2.00 s and t 3.00 s. Answer f i t 2.00 rad/s (3.50 rad/s 2 )(2.00 s) 9.00 rad/s 0.8 rad. TABLE Kineatic Equations for Rotational and Linear Motion Under Constant Acceleration Rotational Motion About a Fied Ais ANGULAR AND LINEAR QUANTITIES In this section we derive soe useful relationships between the angular speed and acceleration of a rotating rigid object and the linear speed and acceleration of an arbitrary point in the object. To do so, we ust keep in ind that when a rigid object rotates about a fied ais, as in Figure 0.4, every particle of the object oves in a circle whose center is the ais of rotation. We can relate the angular speed of the rotating object to the tangential speed of a point P on the object. Because point P oves in a circle, the linear velocity vector v is always tangent to the circular path and hence is called tangential velocity. The agnitude of the tangential velocity of the point P is by definition the tangential speed v ds/dt, where s is the distance traveled by this point easured along the circular path. Recalling that s r (Eq. 0.a) and noting that r is constant, we obtain v ds r d dt dt Because d/dt (see Eq. 0.4), we can say v r (0.0) That is, the tangential speed of a point on a rotating rigid object equals the perpendicular distance of that point fro the ais of rotation ultiplied by the angular speed. Therefore, although every point on the rigid object has the sae angular speed, not every point has the sae linear speed because r is not the sae for all points on the object. Equation 0.0 shows that the linear speed of a point on the rotating object increases as one oves outward fro the center of rotation, as we would intuitively epect. The outer end of a swinging baseball bat oves uch faster than the handle. We can relate the angular acceleration of the rotating rigid object to the tangential acceleration of the point P by taking the tie derivative of v: a t dv r d dt dt Linear Motion f i t v f v i at f i i t t 2 2 f i v i t 2 at 2 2 f 2 i 2( f i ) v 2 f v 2 i 2a( f i ) a t r (0.) That is, the tangential coponent of the linear acceleration of a point on a rotating rigid object equals the point s distance fro the ais of rotation ultiplied by the angular acceleration. y v r θ Figure 0.4 As a rigid object rotates about the fied ais through, the point P has a linear velocity v that is always tangent to the circular path of radius r. Relationship between linear and angular speed QuickLab Spin a tennis ball or basketball and watch it gradually slow down and stop. Estiate and a t as accurately as you can. Relationship between linear and angular acceleration P

27 298 CHAPTER 0 Rotation of a Rigid bject About a Fied Ais 0.4 Rotational Energy 299 y Figure 0.5 As a rigid object rotates about a fied ais through, the point P eperiences a tangential coponent of linear acceleration a t and a radial coponent of linear acceleration a r. The total linear acceleration of this point is a a t a r. EXAMPLE 0.2 CD Player In Section 4.4 we found that a point rotating in a circular path undergoes a centripetal, or radial, acceleration a r of agnitude v 2 /r directed toward the center of rotation (Fig. 0.5). Because v r for a point P on a rotating object, we can epress the radial acceleration of that point as a r v 2 r2 (0.2) r The total linear acceleration vector of the point is a a t a r. (a t describes the change in how fast the point is oving, and a r represents the change in its direction of travel.) Because a is a vector having a radial and a tangential coponent, the agnitude of a for the point P on the rotating rigid object is Quick Quiz 0.2 n a copact disc, audio inforation is stored in a series of pits and flat areas on the surface of the disc. The inforation is stored digitally, and the alternations between pits and flat areas on the surface represent binary ones and zeroes to be read by the copact disc player and converted back to sound waves. The pits and flat areas are detected by a syste consisting of a laser and lenses. The length of a certain nuber of ones and zeroes is the sae everywhere on the disc, whether the inforation is near the center of the disc or near its outer edge. In order that this length of ones and zeroes always passes by the laser lens syste in the sae tie period, the linear speed of the disc surface at the location of the lens ust be constant. This requires, according to Equation 0.0, that the angular speed vary as the laser lens syste oves radially along the disc. In a typical copact disc player, the disc spins counterclockwise (Fig. 0.6), and the constant speed of the surface at the point of the laser lens syste is.3 /s. (a) Find the angular speed of the disc in revolutions per inute when inforation is being read fro the innerost first track (r 23 ) and the outerost final track (r 58 ). Using Equation 0.0, we can find the angular speed; this will give us the required linear speed at the position of the inner track, i v r i a a t a r P.3 /s rad/s a a 2 t a 2 r r 22 r 24 r 2 4 (0.3) When a wheel of radius R rotates about a fied ais, do all points on the wheel have (a) the sae angular speed and (b) the sae linear speed? If the angular speed is constant and equal to, describe the linear speeds and linear accelerations of the points located at (c) r 0, (d) r R/2, and (e) r R, all easured fro the center of the wheel. (56.5 rad/s) rev/in Figure 0.6 rev/rad (60 s/in) 23 A copact disc For the outer track, The player adjusts the angular speed of the disc within this range so that inforation oves past the objective lens at a constant rate. These angular velocity values are positive because the direction of rotation is counterclockwise. (b) The aiu playing tie of a standard usic CD is 74 inutes and 33 seconds. How any revolutions does the disc ake during that tie? f v r f rev/in We know that the angular speed is always decreasing, and we assue that it is decreasing steadily, with constant. The tie interval t is (74 in)(60 s/in) 33 s s. We are looking for the angular position f, where we set the initial angular position i 0. We can use Equation 0.3, replacing the average angular speed with its atheatical equivalent ( i f )/2: 0.4 f i 2 (i f)t 0 2 (540 rev/in 20 rev/in) ( in/60 s)(4 473 s).3 /s rad/s rev RTATINAL ENERGY (c) What total length of track oves past the objective lens during this tie? Because we know the (constant) linear velocity and the tie interval, this is a straightforward calculation: More than 3.6 iles of track spins past the objective lens! (d) What is the angular acceleration of the CD over the s tie interval? Assue that is constant. f v i t (.3 /s)(4 473 s) We have several choices for approaching this proble. Let us use the ost direct approach by utilizing Equation 0.5, which is based on the definition of the ter we are seeking. We should obtain a negative nuber for the angular acceleration because the disc spins ore and ore slowly in the positive direction as tie goes on. ur answer should also be fairly sall because it takes such a long tie ore than an hour for the change in angular speed to be accoplished: f i t Let us now look at the kinetic energy of a rotating rigid object, considering the object as a collection of particles and assuing it rotates about a fied z ais with an angular speed (Fig. 0.7). Each particle has kinetic energy deterined by its ass and linear speed. If the ass of the ith particle is i and its linear speed is v i, its kinetic energy is To proceed further, we ust recall that although every particle in the rigid object has the sae angular speed, the individual linear speeds depend on the distance r i fro the ais of rotation according to the epression v i r i (see Eq. 0.0). The total kinetic energy of the rotating rigid object is the su of the kinetic energies of the individual particles: K R K i i 2 iv 2 i 2 i i r 22 i i We can write this epression in the for K i 2 2 iv i K R 2 i i r i rad/s rad/s 56.5 rad/s s The disc eperiences a very gradual decrease in its rotation rate, as epected. (0.4) where we have factored 2 fro the su because it is coon to every particle. web If you want to learn ore about the physics of CD players, visit the Special Interest Group on CD Applications and Technology at y r i θ v i i Figure 0.7 A rigid object rotating about a z ais with angular speed. The kinetic energy of the particle of ass i is 2 iv 2 i. The total kinetic energy of the object is called its rotational kinetic energy.

28 300 CHAPTER 0 Rotation of a Rigid bject About a Fied Ais 0.5 Calculation of Moents of Inertia 30 Moent of inertia Rotational kinetic energy EXAMPLE 0.3 We siplify this epression by defining the quantity in parentheses as the oent of inertia I: I i i r i 2 (0.5) Fro the definition of oent of inertia, we see that it has diensions of ML 2 (kg 2 in SI units). With this notation, Equation 0.4 becoes K R 2 I2 (0.6) Although we coonly refer to the quantity 2I 2 as rotational kinetic energy, it is not a new for of energy. It is ordinary kinetic energy because it is derived fro a su over individual kinetic energies of the particles contained in the rigid object. However, the atheatical for of the kinetic energy given by Equation 0.6 is a convenient one when we are dealing with rotational otion, provided we know how to calculate I. It is iportant that you recognize the analogy between kinetic energy associated with linear otion 2 v 2 and rotational kinetic energy 2 I2. The quantities I and in rotational otion are analogous to and v in linear otion, respectively. (In fact, I takes the place of every tie we copare a linear-otion equation with its rotational counterpart.) The oent of inertia is a easure of the resistance of an object to changes in its rotational otion, just as ass is a easure of the tendency of an object to resist changes in its linear otion. Note, however, that ass is an intrinsic property of an object, whereas I depends on the physical arrangeent of that ass. Can you think of a situation in which an object s oent of inertia changes even though its ass does not? The ygen Molecule Consider an oygen olecule ( 2 ) rotating in the y plane about the z ais. The ais passes through the center of the olecule, perpendicular to its length. The ass of each oygen ato is kg, and at roo teperature the average separation between the two atos is d (the atos are treated as point asses). (a) Calculate the oent of inertia of the olecule about the z ais. This is a straightforward application of the definition of I. Because each ato is a distance d/2 fro the z ais, the oent of inertia about the ais is I i r 2 i i d 2 2 d d 2 2 ( kg)( ) 2 This is a very sall nuber, consistent with the inuscule asses and distances involved. (b) If the angular speed of the olecule about the z ais is rad/s, what is its rotational kinetic energy? We apply the result we just calculated for the oent of inertia in the forula for K R : K R 2 I2 2 ( kg 2 )( rad/s) kg J Civil engineers use oent of inertia to characterize the elastic properties (rigidity) of such structures as loaded beas. Hence, it is often useful even in a nonrotational contet. 7.5 EXAMPLE Four Rotating Masses Four tiny spheres are fastened to the corners of a frae of negligible ass lying in the y plane (Fig. 0.8). We shall assue that the spheres radii are sall copared with the diensions of the frae. (a) If the syste rotates about the y ais with an angular speed, find the oent of inertia and the rotational kinetic energy about this ais. First, note that the two spheres of ass, which lie on the y ais, do not contribute to I y (that is, r i 0 for these spheres about this ais). Applying Equation 0.5, we obtain M Figure 0.8 I y i i r i 2 Ma 2 Ma 2 a y 2Ma 2 The four spheres are at a fied separation as shown. The oent of inertia of the syste depends on the ais about which it is evaluated. b b a CALCULATIN F MMENTS F INERTIA We can evaluate the oent of inertia of an etended rigid object by iagining the object divided into any sall volue eleents, each of which has ass. We use the definition I r 2 i i and take the liit of this su as : 0. In i this liit, the su becoes an integral over the whole object: I li r 2 i i r 2 d i :0 i (0.7) It is usually easier to calculate oents of inertia in ters of the volue of the eleents rather than their ass, and we can easily ake that change by using Equation., /V, where is the density of the object and V is its volue. We want this epression in its differential for d/dv because the volues we are dealing with are very sall. Solving for d dv and substituting the result M Therefore, the rotational kinetic energy about the y ais is The fact that the two spheres of ass do not enter into this result akes sense because they have no otion about the ais of rotation; hence, they have no rotational kinetic energy. By siilar logic, we epect the oent of inertia about the ais to be I 2b 2 with a rotational kinetic energy about that ais of K R b 2 2. (b) Suppose the syste rotates in the y plane about an ais through (the z ais). Calculate the oent of inertia and rotational kinetic energy about this ais. K R 2 I y2 2 (2Ma 2 )2 Because r i in Equation 0.5 is the perpendicular distance to the ais of rotation, we obtain I z i i r i 2 Ma 2 Ma 2 b 2 b 2 K R 2 I z 2 2 (2Ma2 2b 2 ) 2 Ma 22 2Ma 2 2b 2 (Ma 2 b 2 )2 Coparing the results for parts (a) and (b), we conclude that the oent of inertia and therefore the rotational kinetic energy associated with a given angular speed depend on the ais of rotation. In part (b), we epect the result to include all four spheres and distances because all four spheres are rotating in the y plane. Furtherore, the fact that the rotational kinetic energy in part (a) is saller than that in part (b) indicates that it would take less effort (work) to set the syste into rotation about the y ais than about the z ais.

29 302 CHAPTER 0 Rotation of a Rigid bject About a Fied Ais 0.5 Calculation of Moents of Inertia 303 into Equation 0.7 gives y y I r 2 dv If the object is hoogeneous, then is constant and the integral can be evaluated for a known geoetry. If is not constant, then its variation with position ust be known to coplete the integration. The density given by /V soeties is referred to as volue density for the obvious reason that it relates to volue. ften we use other ways of epressing density. For instance, when dealing with a sheet of unifor thickness t, we can define a surface density t, which signifies ass per unit area. Finally, when ass is distributed along a unifor rod of cross-sectional area A, we soeties use linear density M/L A, which is the ass per unit length. L d Figure 0.0 A unifor rigid rod of length L. The oent of inertia about the y ais is less than that about the y ais. The latter ais is eained in Eaple 0.8. EXAMPLE 0.5 Unifor Hoop Find the oent of inertia of a unifor hoop of ass M and radius R about an ais perpendicular to the plane of the hoop and passing through its center (Fig. 0.9). All ass eleents d are the sae distance r R fro the ais, and so, applying Equation 0.7, we obtain for the oent of inertia about the z ais through : I z r 2 d R 2 d MR 2 Note that this oent of inertia is the sae as that of a single particle of ass M located a distance R fro the ais of rotation. y d R Figure 0.9 The ass eleents d of a unifor hoop are all the sae distance fro. EXAMPLE 0.7 Unifor Solid Cylinder A unifor solid cylinder has a radius R, ass M, and length L. Calculate its oent of inertia about its central ais (the z ais in Fig. 0.). It is convenient to divide the cylinder into any dr L z r R cylindrical shells, each of which has radius r, thickness dr, and length L, as shown in Figure 0.. The volue dv of each shell is its cross-sectional area ultiplied by its length: dv da L (2r dr)l. If the ass per unit volue is, then the ass of this differential volue eleent is d dv 2rL dr. Substituting this epression for d into Equation 0.7, we obtain I z r 2 d 2L R Because the total volue of the cylinder is R 2 L, we see that M/V M/R 2 L. Substituting this value for into the above result gives () I z 2 MR 2 r 3 dr 2LR 4 0 Quick Quiz 0.3 (a) Based on what you have learned fro Eaple 0.5, what do you epect to find for the oent of inertia of two particles, each of ass M/2, located anywhere on a circle of radius R around the ais of rotation? (b) How about the oent of inertia of four particles, each of ass M/4, again located a distance R fro the rotation ais? Figure 0. cylinder. Calculating I about the z ais for a unifor solid Note that this result does not depend on L, the length of the cylinder. In other words, it applies equally well to a long cylinder and a flat disc. Also note that this is eactly half the value we would epect were all the ass concentrated at the outer edge of the cylinder or disc. (See Eaple 0.5.) EXAMPLE 0.6 Unifor Rigid Rod Calculate the oent of inertia of a unifor rigid rod of length L and ass M (Fig. 0.0) about an ais perpendicular to the rod (the y ais) and passing through its center of ass. The shaded length eleent d has a ass d equal to the ass per unit length ultiplied by d: d d M L d Substituting this epression for d into Equation 0.7, with r, we obtain I y r 2 d L/2 2 M L/2 L d M L/2 2 d L L/2 M L 3 3 L/2 L/2 2 ML 2 Table 0.2 gives the oents of inertia for a nuber of bodies about specific aes. The oents of inertia of rigid bodies with siple geoetry (high syetry) are relatively easy to calculate provided the rotation ais coincides with an ais of syetry. The calculation of oents of inertia about an arbitrary ais can be cubersoe, however, even for a highly syetric object. Fortunately, use of an iportant theore, called the parallel-ais theore, often siplifies the calculation. Suppose the oent of inertia about an ais through the center of ass of an object is I CM. The parallel-ais theore states that the oent of inertia about any ais parallel to and a distance D away fro this ais is I I CM MD 2 (0.8) Parallel-ais theore

30 304 CHAPTER 0 Rotation of a Rigid bject About a Fied Ais 0.5 Calculation of Moents of Inertia 305 TABLE 0.2 Moents of Inertia of Hoogeneous Rigid Bodies with Different Geoetries y d, y z Hoop or cylindrical shell I CM = MR 2 R Hollow cylinder I CM = 2 M(R 2 + R 2 2 ) R R 2 y y r CM Rotation ais Ais through CM y y CM D CM, y CM CM CM Solid cylinder or disk I CM = 2 MR2 R Rectangular plate I CM = 2 M(a 2 + b 2 ) Long thin rod with rotation ais through center I CM = 2 ML2 L Solid sphere I CM = 2 5 MR 2 R Long thin rod with rotation ais through end I = 3 ML2 Thin spherical shell I CM = 2 3 MR 2 a b L R (a) Figure 0.2 (a) The parallel-ais theore: If the oent of inertia about an ais perpendicular to the figure through the center of ass is I CM, then the oent of inertia about the z ais is I z I CM MD 2. (b) Perspective drawing showing the z ais (the ais of rotation) and the parallel ais through the CM. and y y y CM. Therefore, I [( CM ) 2 (y y CM ) 2 ] d [() 2 (y) 2 ] d 2 CM d 2y CM y d ( 2 CM y 2 CM ) d The first integral is, by definition, the oent of inertia about an ais that is parallel to the z ais and passes through the center of ass. The second two integrals are zero because, by definition of the center of ass, d y d 0. The last integral is siply MD 2 because d M and D 2 2 CM y 2 CM. Therefore, we conclude that (b) I I CM MD 2 Proof of the Parallel-Ais Theore (ptional). Suppose that an object rotates in the y plane about the z ais, as shown in Figure 0.2, and that the coordinates of the center of ass are CM, y CM. Let the ass eleent d have coordinates, y. Because this eleent is a distance r 2 y 2 fro the z ais, the oent of inertia about the z ais is I r 2 d ( 2 y 2 ) d However, we can relate the coordinates, y of the ass eleent d to the coordinates of this sae eleent located in a coordinate syste having the object s center of ass as its origin. If the coordinates of the center of ass are CM, y CM in the original coordinate syste centered on, then fro Figure 0.2a we see that the relationships between the unpried and pried coordinates are CM EXAMPLE 0.8 Applying the Parallel-Ais Theore Consider once again the unifor rigid rod of ass M and length L shown in Figure 0.0. Find the oent of inertia of the rod about an ais perpendicular to the rod through one end (the yais in Fig. 0.0). Intuitively, we epect the oent of inertia to be greater than I CM 2 ML 2 because it should be ore difficult to change the rotational otion of a rod spinning about an ais at one end than one that is spinning about its center. Because the distance between the center-of-ass ais and the y ais is D L/2, the parallel-ais theore gives I I CM MD 2 2 ML 2 M L 2 2 So, it is four ties ore difficult to change the rotation of a rod spinning about its end than it is to change the otion of one spinning about its center. Eercise Calculate the oent of inertia of the rod about a perpendicular ais through the point L/4. Answer I 7 48 ML 2. 3 ML 2

31 306 CHAPTER 0 Rotation of a Rigid bject About a Fied Ais 0.7 Relationship Between Torque and Angular Acceleration 307 Definition of torque Moent ar TRQUE Why are a door s doorknob and hinges placed near opposite edges of the door? This question actually has an answer based on coon sense ideas. The harder we push against the door and the farther we are fro the hinges, the ore likely we are to open or close the door. When a force is eerted on a rigid object pivoted about an ais, the object tends to rotate about that ais. The tendency of a force to rotate an object about soe ais is easured by a vector quantity called torque (tau). Consider the wrench pivoted on the ais through in Figure 0.3. The applied force F acts at an angle to the horizontal. We define the agnitude of the torque associated with the force F by the epression (0.9) where r is the distance between the pivot point and the point of application of F and d is the perpendicular distance fro the pivot point to the line of action of F. (The line of action of a force is an iaginary line etending out both ends of the vector representing the force. The dashed line etending fro the tail of F in Figure 0.3 is part of the line of action of F.) Fro the right triangle in Figure 0.3 that has the wrench as its hypotenuse, we see that d r sin. This quantity d is called the oent ar (or lever ar) of F. It is very iportant that you recognize that torque is defined only when a reference ais is specified. Torque is the product of a force and the oent ar of that force, and oent ar is defined only in ters of an ais of rotation. In Figure 0.3, the only coponent of F that tends to cause rotation is F sin, the coponent perpendicular to r. The horizontal coponent F cos, because it passes through, has no tendency to produce rotation. Fro the definition of torque, we see that the rotating tendency increases as F increases and as d increases. This eplains the observation that it is easier to close a door if we push at the doorknob rather than at a point close to the hinge. We also want to apply our push as close to perpendicular to the door as we can. Pushing sideways on the doorknob will not cause the door to rotate. If two or ore forces are acting on a rigid object, as shown in Figure 0.4, each tends to produce rotation about the pivot at. In this eaple, F 2 tends to rf sin Fd rotate the object clockwise, and F tends to rotate it counterclockwise. We use the convention that the sign of the torque resulting fro a force is positive if the turning tendency of the force is counterclockwise and is negative if the turning tendency is clockwise. For eaple, in Figure 0.4, the torque resulting fro F, which has a oent ar d, is positive and equal to F d ; the torque fro F 2 is negative and equal to F 2 d 2. Hence, the net torque about is Torque should not be confused with force. Forces can cause a change in linear otion, as described by Newton s second law. Forces can also cause a change in rotational otion, but the effectiveness of the forces in causing this change depends on both the forces and the oent ars of the forces, in the cobination that we call torque. Torque has units of force ties length newton eters in SI units and should be reported in these units. Do not confuse torque and work, which have the sae units but are very different concepts. EXAMPLE F d F 2 d 2 The Net Torque on a Cylinder A one-piece cylinder is shaped as shown in Figure 0.5, with a core section protruding fro the larger dru. The cylinder is free to rotate around the central ais shown in the drawing. A rope wrapped around the dru, which has radius R, eerts a force F to the right on the cylinder. A rope wrapped around the core, which has radius R 2, eerts a force F 2 downward on the cylinder. (a) What is the net torque acting on the cylinder about the rotation ais (which is the z ais in Figure 0.5)? y R 2 R F The torque due to F is R F (the sign is negative because the torque tends to produce clockwise rotation). The torque due to F 2 is R 2 F 2 (the sign is positive because the torque tends to produce counterclockwise rotation). Therefore, the net torque about the rotation ais is 2 R F R 2 F 2 We can ake a quick check by noting that if the two forces are of equal agnitude, the net torque is negative because R R 2. Starting fro rest with both forces acting on it, the cylinder would rotate clockwise because F would be ore effective at turning it than would F 2. (b) Suppose F 5.0 N, R.0, F N, and R What is the net torque about the rotation ais, and which way does the cylinder rotate fro rest? (5.0 N)(.0 ) (5.0 N)(0.50 ) 2.5 N d r F sin φ F φ φ F cos φ Line of action Figure 0.3 The force F has a greater rotating tendency about as F increases and as the oent ar d increases. It is the coponent F sin that tends to rotate the wrench about. d 2 d F 2 F Figure 0.4 The force F tends to rotate the object counterclockwise about, and F 2 tends to rotate it clockwise z F 2 Figure 0.5 A solid cylinder pivoted about the z ais through. The oent ar of F is R, and the oent ar of F 2 is R 2. RELATINSHIP BETWEEN TRQUE AND ANGULAR ACCELERATIN In this section we show that the angular acceleration of a rigid object rotating about a fied ais is proportional to the net torque acting about that ais. Before discussing the ore cople case of rigid-body rotation, however, it is instructive Because the net torque is positive, if the cylinder starts fro rest, it will coence rotating counterclockwise with increasing angular velocity. (If the cylinder s initial rotation is clockwise, it will slow to a stop and then rotate counterclockwise with increasing angular speed.)

32 308 CHAPTER 0 Rotation of a Rigid bject About a Fied Ais 0.7 Relationship Between Torque and Angular Acceleration 309 first to discuss the case of a particle rotating about soe fied point under the influence of an eternal force. Consider a particle of ass rotating in a circle of radius r under the influence of a tangential force F t and a radial force F r, as shown in Figure 0.6. (As we learned in Chapter 6, the radial force ust be present to keep the particle oving in its circular path.) The tangential force provides a tangential acceleration a t, and portional to the angular acceleration of the object, with the proportionality factor being I, a quantity that depends upon the ais of rotation and upon the size and shape of the object. In view of the cople nature of the syste, it is interesting to note that the relationship I is strikingly siple and in coplete agreeent with eperiental observations. The siplicity of the result lies in the anner in which the otion is described. Figure 0.6 A particle rotating in a circle under the influence of a tangential force F t. A force F r in the radial direction also ust be present to aintain the circular otion. Relationship between torque and angular acceleration y d F t r d Figure 0.7 A rigid object rotating about an ais through. Each ass eleent d rotates about with the sae angular acceleration, and the net torque on the object is proportional to. Torque is proportional to angular acceleration The torque about the center of the circle due to F t is Because the tangential acceleration is related to the angular acceleration through the relationship a t r (see Eq. 0.), the torque can be epressed as Recall fro Equation 0.5 that r 2 is the oent of inertia of the rotating particle about the z ais passing through the origin, so that (0.20) That is, the torque acting on the particle is proportional to its angular acceleration, and the proportionality constant is the oent of inertia. It is iportant to note that is the rotational analog of Newton s second law of otion, F a. Now let us etend this discussion to a rigid object of arbitrary shape rotating about a fied ais, as shown in Figure 0.7. The object can be regarded as an infinite nuber of ass eleents d of infinitesial size. If we ipose a cartesian coordinate syste on the object, then each ass eleent rotates in a circle about the origin, and each has a tangential acceleration a t produced by an eternal tangential force df t. For any given eleent, we know fro Newton s second law that I F t a t F t r (a t )r (r)r (r 2 ) I df t (d)a t The torque d associated with the force df t acts about the origin and is given by d r df t (r d)a t Because a t r, the epression for d becoes d (r d)r (r 2 d) It is iportant to recognize that although each ass eleent of the rigid object ay have a different linear acceleration a t, they all have the sae angular acceleration. With this in ind, we can integrate the above epression to obtain the net torque about due to the eternal forces: (r 2 d) r 2 d where can be taken outside the integral because it is coon to all ass eleents. Fro Equation 0.7, we know that r 2 d is the oent of inertia of the object about the rotation ais through, and so the epression for becoes (0.2) Note that this is the sae relationship we found for a particle rotating in a circle (see Eq. 0.20). So, again we see that the net torque about the rotation ais is pro- I Although each point on a rigid object rotating about a fied ais ay not eperience the sae force, linear acceleration, or linear speed, each point eperiences the sae angular acceleration and angular speed at any instant. Therefore, at any instant the rotating rigid object as a whole is characterized by specific values for angular acceleration, net torque, and angular speed. Finally, note that the result I also applies when the forces acting on the ass eleents have radial coponents as well as tangential coponents. This is because the line of action of all radial coponents ust pass through the ais of rotation, and hence all radial coponents produce zero torque about that ais. EXAMPLE 0.0 A unifor rod of length L and ass M is attached at one end to a frictionless pivot and is free to rotate about the pivot in the vertical plane, as shown in Figure 0.8. The rod is released fro rest in the horizontal position. What is the initial angular acceleration of the rod and the initial linear acceleration of its right end? We cannot use our kineatic equations to find or a because the torque eerted on the rod varies with its position, and so neither acceleration is constant. We have enough inforation to find the torque, however, which we can then use in the torque angular acceleration relationship (Eq. 0.2) to find and then a. The only force contributing to torque about an ais through the pivot is the gravitational force Mg eerted on the rod. (The force eerted by the pivot on the rod has zero torque about the pivot because its oent ar is zero.) To Figure 0.8 Pivot L/2 Rotating Rod Mg The unifor rod is pivoted at the left end. copute the torque on the rod, we can assue that the gravitational force acts at the center of ass of the rod, as shown in Figure 0.8. The torque due to this force about an ais through the pivot is With I, and I 3 ML 2 for this ais of rotation (see Table 0.2), we obtain I g L g (L/2) 3 L 2 All points on the rod have this angular acceleration. To find the linear acceleration of the right end of the rod, we use the relationship a t r (Eq. 0.), with r L: a t L Every point has the sae and QuickLab Tip over a child s tall tower of blocks. Try this several ties. Does the tower break at the sae place each tie? What affects where the tower coes apart as it tips? If the tower were ade of toy bricks that snap together, what would happen? (Refer to Conceptual Eaple 0..) 3 2 g 3g 2L This result that a t g for the free end of the rod is rather interesting. It eans that if we place a coin at the tip of the rod, hold the rod in the horizontal position, and then release the rod, the tip of the rod falls faster than the coin does! ther points on the rod have a linear acceleration that 3 is less than 2 g. For eaple, the iddle of the rod has an acceleration of 3 4 g. 2

33 30 CHAPTER 0 Rotation of a Rigid bject About a Fied Ais 0.7 Relationship Between Torque and Angular Acceleration 3 CNCEPTUAL EXAMPLE 0. When a tall sokestack falls over, it often breaks soewhere along its length before it hits the ground, as shown in Figure 0.9. The sae thing happens with a tall tower of children s toy blocks. Why does this happen? As the sokestack rotates around its base, each higher portion of the sokestack falls with an increasing tangential acceleration. (The tangential acceleration of a given point on the sokestack is proportional to the distance of that portion fro the base.) As the acceleration increases, higher portions of the sokestack eperience an acceleration greater than that which could result fro gravity alone; this is siilar to the situation described in Eaple 0.0. This can happen only if these portions are being pulled downward by a force in addition to the gravitational force. The force that causes this to occur is the shear force fro lower portions of the sokestack. Eventually the shear force that provides this acceleration is greater than the sokestack can withstand, and the sokestack breaks. EXAMPLE 0.2 Angular Acceleration of a Wheel A wheel of radius R, ass M, and oent of inertia I is ounted on a frictionless, horizontal ale, as shown in Figure A light cord wrapped around the wheel supports an object of ass. Calculate the angular acceleration of the wheel, the linear acceleration of the object, and the tension in the cord. The torque acting on the wheel about its ais of rotation is TR, where T is the force eerted by the cord on the ri of the wheel. (The gravitational force eerted by the Earth on the wheel and the noral force eerted by the ale on the wheel both pass through the ais of rotation and thus produce no torque.) Because I, we obtain () I TR TR I Now let us apply Newton s second law to the otion of the object, taking the downward direction to be positive: F y g T a g T (2) a Equations () and (2) have three unknowns,, a, and T. Because the object and wheel are connected by a string that does not slip, the linear acceleration of the suspended object is equal to the linear acceleration of a point on the ri of the Falling Sokestacks and Tubling Blocks Figure 0.20 Figure 0.9 T T g A falling sokestack. R The tension in the cord produces a torque about the ale passing through. M wheel. Therefore, the angular acceleration of the wheel and this linear acceleration are related by a R. Using this fact together with Equations () and (2), we obtain EXAMPLE 0.3 Atwood s Machine Revisited Two blocks having asses and 2 are connected to each other by a light cord that passes over two identical, frictionless pulleys, each having a oent of inertia I and radius R, as shown in Figure 0.2a. Find the acceleration of each block and the tensions T, T 2, and T 3 in the cord. (Assue no slipping between cord and pulleys.) (3) (4) a R TR 2 T g R 2 I Substituting Equation (4) into Equation (2), and solving for a and, we find that We shall define the downward direction as positive for and upward as the positive direction for 2. This allows us to represent the acceleration of both asses by a single variable a and also enables us to relate a positive a to a positive (counterclockwise) angular acceleration. Let us write Newton s second law of otion for each block, using the free-body diagras for the two blocks as shown in Figure 0.2b: () g T a (2) T 3 2 g 2 a Net, we ust include the effect of the pulleys on the otion. Free-body diagras for the pulleys are shown in Figure 0.2c. The net torque about the ale for the pulley on the left is (T T 2 )R, while the net torque for the pulley on the right is (T 2 T 3 )R. Using the relation I for each pulley and noting that each pulley has the sae angular acceleration, we obtain (3) (T T 2 )R I (4) (T 2 T 3 )R I We now have four equations with four unknowns: a, T, T 2, and T 3. These can be solved siultaneously. Adding Equations (3) and (4) gives (5) (T T 3 )R 2I Adding Equations () and (2) gives T 3 T g 2 g ( 2 )a I g (6) T T 3 ( 2 )g ( 2 )a Eercise The wheel in Figure 0.20 is a solid disk of M 2.00 kg, R 30.0 c, and I kg 2. The suspended object has a ass of kg. Find the tension in the cord and the angular acceleration of the wheel. Answer 3.27 N; 0.9 rad/s 2. Substituting Equation (6) into Equation (5), we have Because a/r, this epression can be siplified to (7) a a R [( 2 )g ( 2 )a]r 2I ( 2 )g ( 2 )a 2I a g I/R 2 g R I/R ( 2 )g 2 2 I R 2 a R 2 This value of a can then be substituted into Equations () T T 3 T T g + 2 g (a) (b) n T 2 T 2 T p g p g T 3 (c) Figure 0.2 (a) Another look at Atwood s achine. (b) Free-body diagras for the blocks. (c) Free-body diagras for the pulleys, where p g represents the force of gravity acting on each pulley. n 2

34 32 CHAPTER 0 Rotation of a Rigid bject About a Fied Ais 0.8 Work, Power, and Energy in Rotational Motion 33 and (2) to give T and T 3. Finally, T 2 can be found fro Equation (3) or Equation (4). Note that if 2, the acceleration is positive; this eans that the left block accelerates downward, the right block accelerates upward, and both pulleys accelerate counterclockwise. If 2, then all the values are negative and the otions are reversed. If 2, then no acceleration occurs at all. You should copare these results with those found in Eaple 5.9 on page 29. TABLE 0.3 Rotational Motion About a Fied Ais Useful Equations in Rotational and Linear Motion Linear Motion dθ r ds Figure 0.22 A rigid object rotates about an ais through under the action of an eternal force F applied at P. Power delivered to a rigid object P F φ 0.8 WRK, PWER, AND ENERGY IN RTATINAL MTIN In this section, we consider the relationship between the torque acting on a rigid object and its resulting rotational otion in order to generate epressions for the power and a rotational analog to the work kinetic energy theore. Consider the rigid object pivoted at in Figure Suppose a single eternal force F is applied at P, where F lies in the plane of the page. The work done by F as the object rotates through an infinitesial distance ds r d in a tie dt is dw Fds (F sin )r d where F sin is the tangential coponent of F, or, in other words, the coponent of the force along the displaceent. Note that the radial coponent of F does no work because it is perpendicular to the displaceent. Because the agnitude of the torque due to F about is defined as rf sin by Equation 0.9, we can write the work done for the infinitesial rotation as dw (0.22) The rate at which work is being done by F as the object rotates about the fied ais is dw dt d Because dw/dt is the instantaneous power (see Section 7.5) delivered by the force, and because d/dt, this epression reduces to dw (0.23) dt This epression is analogous to Fv in the case of linear otion, and the epression dw d is analogous to dw F d. d dt Angular speed d/dt Linear speed v d/dt Angular acceleration d/dt Linear acceleration a dv/dt Resultant torque I Resultant force F a If f i t If v f v i at constant f i i t t 2 a constant f i v i t 2 at 2 2 f 2 i 2( f i ) v 2 f v 2 i 2a( f i ) Work f W i Rotational kinetic energy K R 2 I2 Power Angular oentu L I Resultant torque dl/dt d Rearranging this epression and noting that d dw, we obtain d dw I d Integrating this epression, we get for the total work done by the net eternal force acting on a rotating syste W f f 2 (0.24) i i where the angular speed changes fro i to f as the angular position changes fro i to f. That is, d Work W f i F d Kinetic energy K 2 v 2 Power Fv Linear oentu p v Resultant force F dp/dt I d 2 If 2 2 Ii the net work done by eternal forces in rotating a syetric rigid object about a fied ais equals the change in the object s rotational energy. Work kinetic energy theore for rotational otion Work and Energy in Rotational Motion In studying linear otion, we found the energy concept and, in particular, the work kinetic energy theore etreely useful in describing the otion of a syste. The energy concept can be equally useful in describing rotational otion. Fro what we learned of linear otion, we epect that when a syetric object rotates about a fied ais, the work done by eternal forces equals the change in the rotational energy. To show that this is in fact the case, let us begin with I. Using the chain rule fro the calculus, we can epress the resultant torque as I I d dt I d d d dt I d d Table 0.3 lists the various equations we have discussed pertaining to rotational otion, together with the analogous epressions for linear otion. The last two equations in Table 0.3, involving angular oentu L, are discussed in Chapter and are included here only for the sake of copleteness. Quick Quiz 0.4 For a hoop lying in the y plane, which of the following requires that ore work be done by an eternal agent to accelerate the hoop fro rest to an angular speed : (a) rotation about the z ais through the center of the hoop, or (b) rotation about an ais parallel to z passing through a point P on the hoop ri?

35 34 CHAPTER 0 Rotation of a Rigid bject About a Fied Ais Suary 35 EXAMPLE 0.4 EXAMPLE 0.5 Rotating Rod Revisited A unifor rod of length L and ass M is free to rotate on a frictionless pin passing through one end (Fig 0.23). The rod is released fro rest in the horizontal position. (a) What is its angular speed when it reaches its lowest position? The question can be answered by considering the echanical energy of the syste. When the rod is horizontal, it has no rotational energy. The potential energy relative to the lowest position of the center of ass of the rod () is MgL/2. When the rod reaches its lowest position, the Figure 0.23 L/2 E f = K R = 2 Iω ω 2 E i = U = MgL/2 A unifor rigid rod pivoted at rotates in a vertical plane under the action of gravity. Connected Cylinders Consider two cylinders having asses and 2, where 2, connected by a string passing over a pulley, as shown in Figure The pulley has a radius R and oent of h R 2 Figure 0.24 h energy is entirely rotational energy, 2 I2, where I is the oent of inertia about the pivot. Because I 3 ML 2 (see Table 0.2) and because echanical energy is constant, we have E i E f or (b) Deterine the linear speed of the center of ass and the linear speed of the lowest point on the rod when it is in the vertical position. These two values can be deterined fro the relationship between linear and angular speeds. We know fro part (a), and so the linear speed of the center of ass is Because r for the lowest point on the rod is twice what it is for the center of ass, the lowest point has a linear speed equal to inertia I about its ais of rotation. The string does not slip on the pulley, and the syste is released fro rest. Find the linear speeds of the cylinders after cylinder 2 descends through a distance h, and the angular speed of the pulley at this tie. 2 MgL 2 I2 2 ( 3 ML 2 )2 v CM r L 2 2v CM 3gL We are now able to account for the effect of a assive pulley. Because the string does not slip, the pulley rotates. We neglect friction in the ale about which the pulley rotates for the following reason: Because the ale s radius is sall relative to that of the pulley, the frictional torque is uch saller than the torque applied by the two cylinders, provided that their asses are quite different. Mechanical energy is constant; hence, the increase in the syste s kinetic energy (the syste being the two cylinders, the pulley, and the Earth) equals the decrease in its potential energy. Because K i 0 (the syste is initially at rest), we have K K f K i ( 2 v f 2 2 2v f 2 2 If 2 ) 0 where v f is the sae for both blocks. Because v f R f, this epression becoes K 2 2 3g L 2 3gL I R 2v 2 f Fro Figure 0.24, we see that the syste loses potential energy as cylinder 2 descends and gains potential energy as cylinder rises. That is, U 2 2 gh and U gh. Applying the principle of conservation of energy in the for K U U 2 0 gives 2 2 v f I R 2v f 2 gh 2 gh 0 2( 2 )gh /2 2 I 2 R SUMMARY If a particle rotates in a circle of radius r through an angle (easured in radians), the arc length it oves through is s r. The angular displaceent of a particle rotating in a circle or of a rigid object rotating about a fied ais is f i (0.2) The instantaneous angular speed of a particle rotating in a circle or of a rigid object rotating about a fied ais is d dt The instantaneous angular acceleration of a rotating object is d (0.4) (0.6) dt When a rigid object rotates about a fied ais, every part of the object has the sae angular speed and the sae angular acceleration. If a particle or object rotates about a fied ais under constant angular acceleration, one can apply equations of kineatics that are analogous to those for linear otion under constant linear acceleration: f i t (0.7) f i it 2t 2 f 2 i 2 2(f i) (0.8) (0.9) A useful technique in solving probles dealing with rotation is to visualize a linear version of the sae proble. When a rigid object rotates about a fied ais, the angular position, angular speed, and angular acceleration are related to the linear position, linear speed, and linear acceleration through the relationships s r u (0.a) v r (0.0) a t r (0.) Because v f Rf, the angular speed of the pulley at this instant is f v f R 2( 2 )gh R 2 I R 2 Eercise Repeat the calculation of v f, using I applied to the pulley and Newton s second law applied to the two cylinders. Use the procedures presented in Eaples 0.2 and 0.3. /2

36 36 CHAPTER 0 Rotation of a Rigid bject About a Fied Ais Probles 37 You ust be able to easily alternate between the linear and rotational variables that describe a given situation. The oent of inertia of a syste of particles is I i r 2 i i (0.5) If a rigid object rotates about a fied ais with angular speed, its rotational energy can be written K R 2 I2 (0.6) where I is the oent of inertia about the ais of rotation. The oent of inertia of a rigid object is I r 2 d (0.7) 9. Using the results fro Eaple 0.2, how would you calculate the angular speed of the wheel and the linear speed of the suspended ass at t 2 s, if the syste is released fro rest at t 0? Is the epression v R valid in this situation? 0. If a sall sphere of ass M were placed at the end of the rod in Figure 0.23, would the result for be greater than, less than, or equal to the value obtained in Eaple 0.4?. Eplain why changing the ais of rotation of an object changes its oent of inertia. 2. Is it possible to change the translational kinetic energy of an object without changing its rotational energy? 3. Two cylinders having the sae diensions are set into rotation about their long aes with the sae angular speed. ne is hollow, and the other is filled with water. Which cylinder will be easier to stop rotating? Eplain your answer. 4. Must an object be rotating to have a nonzero oent of inertia? 5. If you see an object rotating, is there necessarily a net torque acting on it? 6. Can a (oentarily) stationary object have a nonzero angular acceleration? 7. The polar diaeter of the Earth is slightly less than the equatorial diaeter. How would the oent of inertia of the Earth change if soe ass fro near the equator were reoved and transferred to the polar regions to ake the Earth a perfect sphere? QUESTINS. What is the angular speed of the second hand of a clock? What is the direction of as you view a clock hanging vertically? What is the agnitude of the angular acceleration vector of the second hand? 2. A wheel rotates counterclockwise in the y plane. What is the direction of? What is the direction of if the angular velocity is decreasing in tie? 3. Are the kineatic epressions for,, and valid when the angular displaceent is easured in degrees instead of in radians? 4. A turntable rotates at a constant rate of 45 rev/in. What is its angular speed in radians per second? What is the agnitude of its angular acceleration? 5. Suppose a b and M for the syste of particles described in Figure 0.8. About what ais (, y, or z) does where r is the distance fro the ass eleent d to the ais of rotation. The agnitude of the torque associated with a force F acting on an object is (0.9) where d is the oent ar of the force, which is the perpendicular distance fro soe origin to the line of action of the force. Torque is a easure of the tendency of the force to change the rotation of the object about soe ais. If a rigid object free to rotate about a fied ais has a net eternal torque acting on it, the object undergoes an angular acceleration, where (0.2) The rate at which work is done by an eternal force in rotating a rigid object about a fied ais, or the power delivered, is (0.23) The net work done by eternal forces in rotating a rigid object about a fied ais equals the change in the rotational kinetic energy of the object: W 2 If 2 2 Ii 2 (0.24) Fd I the oent of inertia have the sallest value? the largest value? 6. Suppose the rod in Figure 0.0 has a nonunifor ass distribution. In general, would the oent of inertia about the y ais still equal ML 2 /2? If not, could the oent of inertia be calculated without knowledge of the anner in which the ass is distributed? 7. Suppose that only two eternal forces act on a rigid body, and the two forces are equal in agnitude but opposite in direction. Under what condition does the body rotate? 8. Eplain how you ight use the apparatus described in Eaple 0.2 to deterine the oent of inertia of the wheel. (If the wheel does not have a unifor ass density, the oent of inertia is not necessarily equal to 2 MR 2.) PRBLEMS, 2, 3 = straightforward, interediate, challenging = full solution available in the Student s Manual and Study Guide WEB = solution posted at = Coputer useful in solving proble = Interactive Physics = paired nuerical/sybolic probles Section 0.2 Rotational Kineatics: Rotational Motion with Constant Angular Acceleration. A wheel starts fro rest and rotates with constant angular acceleration and reaches an angular speed of 2.0 rad/s in 3.00 s. Find (a) the agnitude of the angular acceleration of the wheel and (b) the angle (in radians) through which it rotates in this tie. 2. What is the angular speed in radians per second of (a) the Earth in its orbit about the Sun and (b) the Moon in its orbit about the Earth? 3. An airliner arrives at the terinal, and its engines are shut off. The rotor of one of the engines has an initial clockwise angular speed of rad/s. The engine s rotation slows with an angular acceleration of agnitude 80.0 rad/s 2. (a) Deterine the angular speed after 0.0 s. (b) How long does it take for the rotor to coe to rest? 4. (a) The positions of the hour and inute hand on a clock face coincide at 2 o clock. Deterine all other ties (up to the second) at which the positions of the hands coincide. (b) If the clock also has a second hand, deterine all ties at which the positions of all three hands coincide, given that they all coincide at 2 o clock. 5. An electric otor rotating a grinding wheel at 00 rev/in is switched off. Assuing constant negative acceleration of agnitude 2.00 rad/s 2, (a) how long does it take the wheel to stop? (b) Through how any radians does it turn during the tie found in part (a)? 6. A centrifuge in a edical laboratory rotates at a rotational speed of rev/in. When switched off, it rotates 50.0 ties before coing to rest. Find the constant angular acceleration of the centrifuge. WEB 7. The angular position of a swinging door is described by t 2.00t 2 rad. Deterine the angular position, angular speed, and angular acceleration of the door (a) at t 0 and (b) at t 3.00 s. 8. The tub of a washer goes into its spin cycle, starting fro rest and gaining angular speed steadily for 8.00 s, when it is turning at 5.00 rev/s. At this point the person doing the laundry opens the lid, and a safety switch turns off the washer. The tub soothly slows to rest in 2.0 s. Through how any revolutions does the tub turn while it is in otion? 9. A rotating wheel requires 3.00 s to coplete 37.0 revolutions. Its angular speed at the end of the 3.00-s interval is 98.0 rad/s. What is the constant angular acceleration of the wheel? 0. (a) Find the angular speed of the Earth s rotation on its ais. As the Earth turns toward the east, we see the sky turning toward the west at this sae rate. (b) The rainy Pleiads wester And seek beyond the sea The head that I shall drea of That shall not drea of e. A. E. Housan ( Robert E. Syons) Cabridge, England, is at longitude 0, and Saskatoon, Saskatchewan, is at longitude 07 west. How uch tie elapses after the Pleiades set in Cabridge until these stars fall below the western horizon in Saskatoon? Section 0.3 Angular and Linear Quantities. Make an order-of-agnitude estiate of the nuber of revolutions through which a typical autoobile tire

37 38 CHAPTER 0 Rotation of a Rigid bject About a Fied Ais Probles 39 turns in yr. State the quantities you easure or estiate and their values. 2. The diaeters of the ain rotor and tail rotor of a single-engine helicopter are 7.60 and.02, respectively. The respective rotational speeds are 450 rev/in and 4 38 rev/in. Calculate the speeds of the tips of both rotors. Copare these speeds with the speed of sound, 343 /s. sue the discus oves on the arc of a circle.00 in radius. (a) Calculate the final angular speed of the discus. (b) Deterine the agnitude of the angular acceleration of the discus, assuing it to be constant. (c) Calculate the acceleration tie. 7. A car accelerates uniforly fro rest and reaches a speed of 22.0 /s in 9.00 s. If the diaeter of a tire is 58.0 c, find (a) the nuber of revolutions the tire akes during this otion, assuing that no slipping occurs. (b) What is the final rotational speed of a tire in revolutions per second? 8. A 6.00-kg block is released fro A on the frictionless track shown in Figure P0.8. Deterine the radial and tangential coponents of acceleration for the block at P. 22. A standard cassette tape is placed in a standard cassette player. Each side plays for 30 in. The two tape wheels of the cassette fit onto two spindles in the player. Suppose that a otor drives one spindle at a constant angular speed of rad/s and that the other spindle is free to rotate at any angular speed. Estiate the order of agnitude of the thickness of the tape. Section 0.4 Rotational Energy 23. Three sall particles are connected by rigid rods of negligible ass lying along the y ais (Fig. P0.23). If the syste rotates about the ais with an angular speed of 2.00 rad/s, find (a) the oent of inertia about the ais and the total rotational kinetic energy evaluated fro 2 I2 and (b) the linear speed of each particle and the total kinetic energy evaluated fro 2 2 iv i. y() 3.00 kg 2.00 kg kg 4.00 kg Figure P0.25 () A y Figure P0.2 (Ross Harrrison Koty/Tony Stone Iages) h = 5.00 P 4.00 kg y = A racing car travels on a circular track with a radius of 250. If the car oves with a constant linear speed of 45.0 /s, find (a) its angular speed and (b) the agnitude and direction of its acceleration. 4. A car is traveling at 36.0 k/h on a straight road. The radius of its tires is 25.0 c. Find the angular speed of one of the tires, with its ale taken as the ais of rotation. 5. A wheel 2.00 in diaeter lies in a vertical plane and rotates with a constant angular acceleration of 4.00 rad/s 2. The wheel starts at rest at t 0, and the radius vector of point P on the ri akes an angle of 57.3 with the horizontal at this tie. At t 2.00 s, find (a) the angular speed of the wheel, (b) the linear speed and acceleration of the point P, and (c) the angular position of the point P. 6. A discus thrower accelerates a discus fro rest to a speed of 25.0 /s by whirling it through.25 rev. As- Figure P0.6 (Bruce Ayers/Tony Stone Iages) WEB Figure P0.8 R = A disc 8.00 c in radius rotates at a constant rate of 200 rev/in about its central ais. Deterine (a) its angular speed, (b) the linear speed at a point 3.00 c fro its center, (c) the radial acceleration of a point on the ri, and (d) the total distance a point on the ri oves in 2.00 s. 20. A car traveling on a flat (unbanked) circular track accelerates uniforly fro rest with a tangential acceleration of.70 /s 2. The car akes it one quarter of the way around the circle before it skids off the track. Deterine the coefficient of static friction between the car and track fro these data. 2. A sall object with ass 4.00 kg oves counterclockwise with constant speed 4.50 /s in a circle of radius 3.00 centered at the origin. (a) It started at the point with cartesian coordinates (3, 0). When its angular displaceent is 9.00 rad, what is its position vector, in cartesian unit-vector notation? (b) In what quadrant is the particle located, and what angle does its position vector ake with the positive ais? (c) What is its velocity vector, in unit vector notation? (d) In what direction is it oving? Make a sketch of the position and velocity vectors. (e) What is its acceleration, epressed in unit vector notation? (f) What total force acts on the object? (Epress your answer in unit vector notation.) WEB 2.00 kg 3.00 kg y = 2.00 y = 4.00 Figure P The center of ass of a pitched baseball (3.80-c radius) oves at 38.0 /s. The ball spins about an ais through its center of ass with an angular speed of 25 rad/s. Calculate the ratio of the rotational energy to the translational kinetic energy. Treat the ball as a unifor sphere. 25. The four particles in Figure P0.25 are connected by rigid rods of negligible ass. The origin is at the center of the rectangle. If the syste rotates in the y plane about the z ais with an angular speed of 6.00 rad/s, calculate (a) the oent of inertia of the syste about the z ais and (b) the rotational energy of the syste. 26. The hour hand and the inute hand of Big Ben, the faous Parliaent tower clock in London, are 2.70 long and 4.50 long and have asses of 60.0 kg and 00 kg, respectively. Calculate the total rotational kinetic energy of the two hands about the ais of rotation. (You ay odel the hands as long thin rods.) Figure P Two asses M and are connected by a rigid rod of length L and of negligible ass, as shown in Figure P0.27. For an ais perpendicular to the rod, show that the syste has the iniu oent of inertia when the ais passes through the center of ass. Show that this oent of inertia is I L 2, where M/( M). Section 0.5 Probles 26 and 74. ( John Lawrence/Tony Stone Iages) M Figure P0.27 Calculation of Moents of Inertia 28. Three identical thin rods, each of length L and ass, are welded perpendicular to each other, as shown in Figure P0.28. The entire setup is rotated about an ais L L

38 320 CHAPTER 0 Rotation of a Rigid bject About a Fied Ais Probles Attention! About face! Copute an order-of-agnitude estiate for the oent of inertia of your body as you stand tall and turn around a vertical ais passing through the top of your head and the point halfway between your ankles. In your solution state the quantities you easure or estiate and their values N I, R 2 Section 0.6 Torque 32. Find the ass needed to balance the 500-kg truck on the incline shown in Figure P0.32. Assue all pulleys are frictionless and assless. Figure P0.39 θ Figure P0.28 that passes through the end of one rod and is parallel to another. Deterine the oent of inertia of this arrangeent. 29. Figure P0.29 shows a side view of a car tire and its radial diensions. The rubber tire has two sidewalls of unifor thickness c and a tread wall of unifor thickness 2.50 c and width 20.0 c. Suppose its density is unifor, with the value kg/ 3. Find its oent of inertia about an ais through its center perpendicular to the plane of the sidewalls c 30.5 c Ais of rotation 6.5 c Figure P0.29 Sidewall Tread 30. Use the parallel-ais theore and Table 0.2 to find the oents of inertia of (a) a solid cylinder about an ais parallel to the center-of-ass ais and passing through the edge of the cylinder and (b) a solid sphere about an ais tangent to its surface. WEB 500 kg 33. Find the net torque on the wheel in Figure P0.33 about the ale through if a 0.0 c and b 25.0 c. 2.0 N θ = 45 Figure P a 0.0 N 34. The fishing pole in Figure P0.34 akes an angle of 20.0 with the horizontal. What is the torque eerted by b Figure P0.33 r 9.00 N 3r WEB Figure P0.34 the fish about an ais perpendicular to the page and passing through the fisher s hand? 35. The tires of a 500-kg car are in diaeter, and the coefficients of friction with the road surface are s and k Assuing that the weight is evenly distributed on the four wheels, calculate the aiu torque that can be eerted by the engine on a driving wheel such that the wheel does not spin. If you wish, you ay suppose that the car is at rest. 36. Suppose that the car in Proble 35 has a disk brake syste. Each wheel is slowed by the frictional force between a single brake pad and the disk-shaped rotor. n this particular car, the brake pad coes into contact with the rotor at an average distance of 22.0 c fro the ais. The coefficients of friction between the brake pad and the disk are s and k Calculate the noral force that ust be applied to the rotor such that the car slows as quickly as possible. Section 0.7 Relationship Between Torque and Angular Acceleration 37. A odel airplane having a ass of kg is tethered by a wire so that it flies in a circle 30.0 in radius. The airplane engine provides a net thrust of N perpendicular to the tethering wire. (a) Find the torque the net thrust produces about the center of the circle. (b) Find the angular acceleration of the airplane when it is in level flight. (c) Find the linear acceleration of the airplane tangent to its flight path. 38. The cobination of an applied force and a frictional force produces a constant total torque of 36.0 N on a wheel rotating about a fied ais. The applied force acts for 6.00 s; during this tie the angular speed of the wheel increases fro 0 to 0.0 rad/s. The applied force is then reoved, and the wheel coes to rest in 60.0 s. Find (a) the oent of inertia of the wheel, (b) the agnitude of the frictional torque, and (c) the total nuber of revolutions of the wheel. 39. A block of ass 2.00 kg and a block of ass kg are connected by a assless string over a pulley in the shape of a disk having radius R and ass M 0.0 kg. These blocks are allowed to ove on a fied block wedge of angle 30.0, as shown in Figure P0.39. The coefficient of kinetic friction for both blocks is Draw free-body diagras of both blocks and of the pulley. Deterine (a) the acceleration of the two blocks and (b) the tensions in the string on both sides of the pulley. 40. A potter s wheel a thick stone disk with a radius of and a ass of 00 kg is freely rotating at 50.0 rev/in. The potter can stop the wheel in 6.00 s by pressing a wet rag against the ri and eerting a radially inward force of 70.0 N. Find the effective coefficient of kinetic friction between the wheel and the rag. 4. A bicycle wheel has a diaeter of 64.0 c and a ass of.80 kg. Assue that the wheel is a hoop with all of its ass concentrated on the outside radius. The bicycle is placed on a stationary stand on rollers, and a resistive force of 20 N is applied tangent to the ri of the tire. (a) What force ust be applied by a chain passing over a 9.00-c-diaeter sprocket if the wheel is to attain an acceleration of 4.50 rad/s 2? (b) What force is required if the chain shifts to a 5.60-c-diaeter sprocket? Section 0.8 Work, Power, and Energy in Rotational Motion 42. A cylindrical rod 24.0 c long with a ass of.20 kg and a radius of.50 c has a ball with a diaeter of 8.00 c and a ass of 2.00 kg attached to one end. The arrangeent is originally vertical and stationary, with the ball at the top. The apparatus is free to pivot about the botto end of the rod. (a) After it falls through 90, what is its rotational kinetic energy? (b) What is the angular speed of the rod and ball? (c) What is the linear speed of the ball? (d) How does this copare with the speed if the ball had fallen freely through the sae distance of 28 c? 43. A 5.0-kg ass and a 0.0-kg ass are suspended by a pulley that has a radius of 0.0 c and a ass of 3.00 kg (Fig. P0.43). The cord has a negligible ass and causes the pulley to rotate without slipping. The pulley

39 322 CHAPTER 0 Rotation of a Rigid bject About a Fied Ais Probles 323 rotates without friction. The asses start fro rest 3.00 apart. Treating the pulley as a unifor disk, deterine the speeds of the two asses as they pass each other. 44. A ass and a ass 2 are suspended by a pulley that has a radius R and a ass M (see Fig. P0.43). The cord has a negligible ass and causes the pulley to rotate without slipping. The pulley rotates without friction. The asses start fro rest a distance d apart. Treating the pulley as a unifor disk, deterine the speeds of the two asses as they pass each other. M R M = 3.00 kg R = 0.0 c = 5.0 kg 2 = 0.0 kg Figure P0.43 Probles 43 and 44. Figure P0.47 v. Show that the oent of inertia I of the equipent (including the turntable) is r 2 (2gh/v 2 ). 48. A bus is designed to draw its power fro a rotating flywheel that is brought up to its aiu rate of rotation (3 000 rev/in) by an electric otor. The flywheel is a solid cylinder with a ass of 000 kg and a diaeter of.00. If the bus requires an average power of 0.0 kw, how long does the flywheel rotate? 49. (a) A unifor, solid disk of radius R and ass M is free to rotate on a frictionless pivot through a point on its ri (Fig. P0.49). If the disk is released fro rest in the position shown by the blue circle, what is the speed of its center of ass when the disk reaches the position indicated by the dashed circle? (b) What is the speed of the lowest point on the disk in the dashed position? (c) Repeat part (a), using a unifor hoop. withstand uch shear stress. As the chiney begins to fall, shear forces ust act on the topost sections to accelerate the tangentially so that they can keep up with the rotation of the lower part of the stack. For siplicity, let us odel the chiney as a unifor rod of length pivoted at the lower end. The rod starts at rest in a vertical position (with the frictionless pivot at the botto) and falls over under the influence of gravity. What fraction of the length of the rod has a tangential acceleration greater than g sin, where is the angle the chiney akes with the vertical? eerts on the wheel. (a) How long does the wheel take to reach its final rotational speed of 200 rev/in? (b) Through how any revolutions does it turn while accelerating? 54. The density of the Earth, at any distance r fro its center, is approiately [4.2.6 r/r] 0 3 kg/ 3 where R is the radius of the Earth. Show that this density leads to a oent of inertia I 0.330MR 2 about an ais through the center, where M is the ass of the Earth. 55. A length of light nylon cord is wound around a unifor cylindrical spool of radius and ass.00 kg. The spool is ounted on a frictionless ale and is initially at rest. The cord is pulled fro the spool with a constant acceleration of agnitude 2.50 /s 2. (a) How uch work has been done on the spool when it reaches an angular speed of 8.00 rad/s? (b) Assuing that there is enough cord on the spool, how long does it take the spool to reach this angular speed? (c) Is there enough cord on the spool? 56. A flywheel in the for of a heavy circular disk of diaeter and ass 200 kg is ounted on a frictionless bearing. A otor connected to the flywheel accelerates it fro rest to 000 rev/in. (a) What is the oent of inertia of the flywheel? (b) How uch work is done on it during this acceleration? (c) When the angular speed reaches 000 rev/in, the otor is disengaged. A friction brake is used to slow the rotational rate to 500 rev/in. How uch energy is dissipated as internal energy in the friction brake? 57. A shaft is turning at 65.0 rad/s at tie zero. Thereafter, its angular acceleration is given by 0 rad/s 2 5t rad/s A weight of 50.0 N is attached to the free end of a light string wrapped around a reel with a radius of and a ass of 3.00 kg. The reel is a solid disk, free to rotate in a vertical plane about the horizontal ais passing through its center. The weight is released 6.00 above the floor. (a) Deterine the tension in the string, the acceleration of the ass, and the speed with which the weight hits the floor. (b) Find the speed calculated in part (a), using the principle of conservation of energy. 46. A constant torque of 25.0 N is applied to a grindstone whose oent of inertia is 0.30 kg 2. Using energy principles, find the angular speed after the grindstone has ade 5.0 revolutions. (Neglect friction.) 47. This proble describes one eperiental ethod of deterining the oent of inertia of an irregularly shaped object such as the payload for a satellite. Figure P0.47 shows a ass suspended by a cord wound around a spool of radius r, foring part of a turntable supporting the object. When the ass is released fro rest, it descends through a distance h, acquiring a speed Pivot Figure P A horizontal 800-N erry-go-round is a solid disk of radius.50 and is started fro rest by a constant horizontal force of 50.0 N applied tangentially to the cylinder. Find the kinetic energy of the solid cylinder after 3.00 s. ADDITINAL PRBLEMS 5. Toppling chineys often break apart in id-fall (Fig. P0.5) because the ortar between the bricks cannot R g Figure P0.5 A building deolition site in Baltiore, MD. At the left is a chiney, ostly concealed by the building, that has broken apart on its way down. Copare with Figure 0.9. ( Jerry Wachter/Photo Researchers, Inc.) 52. Review Proble. A iing beater consists of three thin rods: Each is 0.0 c long, diverges fro a central hub, and is separated fro the others by 20. All turn in the sae plane. A ball is attached to the end of each rod. Each ball has a cross-sectional area of 4.00 c 2 and is shaped so that it has a drag coefficient of Calculate the power input required to spin the beater at 000 rev/in (a) in air and (b) in water. 53. A grinding wheel is in the for of a unifor solid disk having a radius of 7.00 c and a ass of 2.00 kg. It starts fro rest and accelerates uniforly under the action of the constant torque of N that the otor where t is the elapsed tie. (a) Find its angular speed at t 3.00 s. (b) How far does it turn in these 3 s? 58. For any given rotational ais, the radius of gyration K of a rigid body is defined by the epression K 2 I/M, where M is the total ass of the body and I is its oent of inertia. Thus, the radius of gyration is equal to the distance between an iaginary point ass M and the ais of rotation such that I for the point ass about that ais is the sae as that for the rigid body. Find the radius of gyration of (a) a solid disk of radius R, (b) a unifor rod of length L, and (c) a solid sphere of radius R, all three of which are rotating about a central ais. 59. A long, unifor rod of length L and ass M is pivoted about a horizontal, frictionless pin passing through one end. The rod is released fro rest in a vertical position, as shown in Figure P0.59. At the instant the rod is horizontal, find (a) its angular speed, (b) the agnitude of its angular acceleration, (c) the and y coponents of the acceleration of its center of ass, and (d) the coponents of the reaction force at the pivot.

40 324 CHAPTER 0 Rotation of a Rigid bject About a Fied Ais Probles 325 h L Pivot y Figure P0.59 Figure P0.60 Probles 60 and A bicycle is turned upside down while its owner repairs a flat tire. A friend spins the other wheel, of radius 0.38, and observes that drops of water fly off tangentially. She easures the height reached by drops oving vertically (Fig. P0.60). A drop that breaks loose fro the tire on one turn rises h 54.0 c above the tangent point. A drop that breaks loose on the net turn rises 5.0 c above the tangent point. The height to which the drops rise decreases because the angular speed of the wheel decreases. Fro this inforation, deterine the agnitude of the average angular acceleration of the wheel. 6. A bicycle is turned upside down while its owner repairs a flat tire. A friend spins the other wheel of radius R and observes that drops of water fly off tangentially. She easures the height reached by drops oving vertically (see Fig. P0.60). A drop that breaks loose fro the tire on one turn rises a distance h above the tangent point. A drop that breaks loose on the net turn rises a distance h 2 h above the tangent point. The height to which the drops rise decreases because the angular speed of the wheel decreases. Fro this inforation, deterine the agnitude of the average angular acceleration of the wheel. 62. The top shown in Figure P0.62 has a oent of inertia of kg 2 and is initially at rest. It is free to rotate about the stationary ais AA. A string, wrapped around a peg along the ais of the top, is pulled in such a anner that a constant tension of 5.57 N is aintained. If the string does not slip while it is unwound fro the peg, what is the angular speed of the top after 80.0 c of string has been pulled off the peg? F 63. A cord is wrapped around a pulley of ass and of radius r. The free end of the cord is connected to a block of ass M. The block starts fro rest and then slides down an incline that akes an angle with the horizontal. The coefficient of kinetic friction between block and incline is. (a) Use energy ethods to show that the block s speed as a function of displaceent d down the incline is v [4gdM( 2M) (sin cos )] /2 (b) Find the agnitude of the acceleration of the block in ters of,, M, g, and. 64. (a) What is the rotational energy of the Earth about its spin ais? The radius of the Earth is k, and its ass is kg. Treat the Earth as a sphere of 2 oent of inertia 5 MR 2. (b) The rotational energy of the Earth is decreasing steadily because of tidal friction. Estiate the change in one day, given that the rotational period increases by about 0 s each year. 65. The speed of a oving bullet can be deterined by allowing the bullet to pass through two rotating paper disks ounted a distance d apart on the sae ale (Fig. P0.65). Fro the angular displaceent of the two A Figure P0.62 A v θ = 3 bullet holes in the disks and the rotational speed of the disks, we can deterine the speed v of the bullet. Find the bullet speed for the following data: d 80 c, 900 rev/in, and A wheel is fored fro a hoop and n equally spaced spokes etending fro the center of the hoop to its ri. The ass of the hoop is M, and the radius of the hoop (and hence the length of each spoke) is R. The ass of each spoke is. Deterine (a) the oent of inertia of the wheel about an ais through its center and perpendicular to the plane of the wheel and (b) the oent of inertia of the wheel about an ais through its ri and perpendicular to the plane of the wheel. 67. A unifor, thin, solid door has a height of 2.20, a width of 0.870, and a ass of 23.0 kg. Find its oent of inertia for rotation on its hinges. Are any of the data unnecessary? 68. A unifor, hollow, cylindrical spool has inside radius R/2, outside radius R, and ass M (Fig. P0.68). It is ounted so that it rotates on a assless horizontal ale. A ass is connected to the end of a string wound around the spool. The ass falls fro rest through a distance y in tie t. Show that the torque due to the frictional forces between spool and ale is f R[(g 2y/t 2 ) M(5y/4t 2 )] 69. An electric otor can accelerate a Ferris wheel of oent of inertia I kg 2 fro rest to R/2 M R/2 ω d Figure P0.65 Figure P0.68 y 0.0 rev/in in 2.0 s. When the otor is turned off, friction causes the wheel to slow down fro 0.0 to 8.00 rev/in in 0.0 s. Deterine (a) the torque generated by the otor to bring the wheel to 0.0 rev/in and (b) the power that would be needed to aintain this rotational speed. 70. The pulley shown in Figure P0.70 has radius R and oent of inertia I. ne end of the ass is connected to a spring of force constant k, and the other end is fastened to a cord wrapped around the pulley. The pulley ale and the incline are frictionless. If the pulley is wound counterclockwise so that the spring is stretched a distance d fro its unstretched position and is then released fro rest, find (a) the angular speed of the pulley when the spring is again unstretched and (b) a nuerical value for the angular speed at this point if I.00 kg 2, R 0.300, k 50.0 N/, kg, d 0.200, and R 7. Two blocks, as shown in Figure P0.7, are connected by a string of negligible ass passing over a pulley of radius and oent of inertia I. The block on the frictionless incline is oving upward with a constant acceleration of 2.00 /s 2. (a) Deterine T and T 2, the tensions in the two parts of the string. (b) Find the oent of inertia of the pulley. 72. A coon deonstration, illustrated in Figure P0.72, consists of a ball resting at one end of a unifor board 5.0 kg 2.00 /s T θ Figure P0.70 Figure P0.7 k 2 T kg

41 326 CHAPTER 0 Rotation of a Rigid bject About a Fied Ais of length, hinged at the other end, and elevated at an angle. A light cup is attached to the board at r c so that it will catch the ball when the support stick is suddenly Hinged end θ r c reoved. (a) Show that the ball will lag behind the falling board when is less than 35.3 ; and that (b) the ball will fall into the cup when the board is supported at Figure P0.72 Cup Support stick this liiting angle and the cup is placed at r c 2 3 cos (c) If a ball is at the end of a.00- stick at this critical angle, show that the cup ust be 8.4 c fro the oving end. 73. As a result of friction, the angular speed of a wheel changes with tie according to the relationship d/dt 0e t where 0 and are constants. The angular speed changes fro 3.50 rad/s at t 0 to 2.00 rad/s at t 9.30 s. Use this inforation to deterine and 0. Then, deterine (a) the agnitude of the angular acceleration at t 3.00 s, (b) the nuber of revolutions the wheel akes in the first 2.50 s, and (c) the nuber of revolutions it akes before coing to rest. 74. The hour hand and the inute hand of Big Ben, the faous Parliaent tower clock in London, are 2.70 long and 4.50 long and have asses of 60.0 kg and 00 kg, respectively (see Fig. P0.26). (a) Deterine the total torque due to the weight of these hands about the ais of rotation when the tie reads (i) 3:00, (ii) 5:5, (iii) 6:00, (iv) 8:20, and (v) 9:45. (You ay odel the hands as long thin rods.) (b) Deterine all ties at which the total torque about the ais of rotation is zero. Deterine the ties to the nearest second, solving a transcendental equation nuerically. ANSWERS T QUICK QUIZZES 0. The fact that is negative indicates that we are dealing with an object that is rotating in the clockwise direction. We also know that when and are antiparallel, ust be decreasing the object is slowing down. Therefore, the object is spinning ore and ore slowly (with less and less angular speed) in the clockwise, or negative, direction. This has a linear analogy to a sky diver opening her parachute. The velocity is negative downward. When the sky diver opens the parachute, a large upward force causes an upward acceleration. As a result, the acceleration and velocity vectors are in opposite directions. Consequently, the parachutist slows down. 0.2 (a) Yes, all points on the wheel have the sae angular speed. This is why we use angular quantities to describe rotational otion. (b) No, not all points on the wheel have the sae linear speed. (c) v 0, a 0. (d) v R/2, a a r v 2 /(R/2) R2 /2 (a t is zero at all points because is constant).(e) v R, a R (a) I MR 2. (b) I MR 2. The oent of inertia of a syste of asses equidistant fro an ais of rotation is always the su of the asses ultiplied by the square of the distance fro the ais. 0.4 (b) Rotation about the ais through point P requires ore work. The oent of inertia of the hoop about the center ais is I CM MR 2, whereas, by the parallelais theore, the oent of inertia about the ais through point P is I P I CM MR 2 MR 2 MR 2 2MR 2.

42 Rolling Motion and Angular Moentu P U Z Z L E R ne of the ost popular early bicycles was the penny farthing, introduced in 870. The bicycle was so naed because the size relationship of its two wheels was about the sae as the size relationship of the penny and the farthing, two English coins. When the rider looks down at the top of the front wheel, he sees it oving forward faster than he and the handlebars are oving. Yet the center of the wheel does not appear to be oving at all relative to the handlebars. How can different parts of the rolling wheel ove at different linear speeds? ( Steve Lovegrove/Tasanian Photo Library) c h a p t e r 328 CHAPTER Rolling Motion and Angular Moentu 7.7 In the preceding chapter we learned how to treat a rigid body rotating about a fied ais; in the present chapter, we ove on to the ore general case in which the ais of rotation is not fied in space. We begin by describing such otion, which is called rolling otion. The central topic of this chapter is, however, angular oentu, a quantity that plays a key role in rotational dynaics. In analogy to the conservation of linear oentu, we find that the angular oentu of a rigid object is always conserved if no eternal torques act on the object. Like the law of conservation of linear oentu, the law of conservation of angular oentu is a fundaental law of physics, equally valid for relativistic and quantu systes.. RLLING MTIN F A RIGID BJECT In this section we treat the otion of a rigid object rotating about a oving ais. In general, such otion is very cople. However, we can siplify atters by restricting our discussion to a hoogeneous rigid object having a high degree of syetry, such as a cylinder, sphere, or hoop. Furtherore, we assue that the object undergoes rolling otion along a flat surface. We shall see that if an object such as a cylinder rolls without slipping on the surface (we call this pure rolling otion), a siple relationship eists between its rotational and translational otions. Suppose a cylinder is rolling on a straight path. As Figure. shows, the center of ass oves in a straight line, but a point on the ri oves in a ore cople path called a cycloid. This eans that the ais of rotation reains parallel to its initial orientation in space. Consider a unifor cylinder of radius R rolling without slipping on a horizontal surface (Fig..2). As the cylinder rotates through an angle, its center of ass oves a linear distance s R (see Eq. 0.a). Therefore, the linear speed of the center of ass for pure rolling otion is given by v CM ds dt R d R dt (.) where is the angular velocity of the cylinder. Equation. holds whenever a cylinder or sphere rolls without slipping and is the condition for pure rolling Chapter utline. Rolling Motion of a Rigid bject.2 The Vector Product and Torque.3 Angular Moentu of a Particle.4 Angular Moentu of a Rotating Rigid bject.5 Conservation of Angular Moentu.6 (ptional) The Motion of Gyroscopes and Tops.7 (ptional) Angular Moentu as a Fundaental Quantity 327 Figure. ne light source at the center of a rolling cylinder and another at one point on the ri illustrate the different paths these two points take. The center oves in a straight line (green line), whereas the point on the ri oves in the path called a cycloid (red curve). (Henry Leap and Ji Lehan)

43 . Rolling Motion of a Rigid bject CHAPTER Rolling Motion and Angular Moentu P v CM P v = Rωω CM v CM CM v = 0 R θ s = Rθ s Figure.2 In pure rolling otion, as the cylinder rotates through an angle, its center of ass oves a linear distance s R. v CM P (a) Pure translation P v = Rωω P (b) Pure rotation v = v CM + Rω ω = 2v CM otion. The agnitude of the linear acceleration of the center of ass for pure rolling otion is CM v = v CM 7.2 a CM dv CM dt R d dt R (.2) where is the angular acceleration of the cylinder. The linear velocities of the center of ass and of various points on and within the cylinder are illustrated in Figure.3. A short tie after the oent shown in the drawing, the ri point labeled P will have rotated fro the si o clock position to, say, the seven o clock position, the point Q will have rotated fro the ten o clock position to the eleven o clock position, and so on. Note that the linear velocity of any point is in a direction perpendicular to the line fro that point to the contact point P. At any instant, the part of the ri that is at point P is at rest relative to the surface because slipping does not occur. All points on the cylinder have the sae angular speed. Therefore, because the distance fro P to P is twice the distance fro P to the center of ass, P has a speed 2v CM 2R. To see why this is so, let us odel the rolling otion of the cylinder in Figure.4 as a cobination of translational (linear) otion and rotational otion. For the pure translational otion shown in Figure.4a, iagine that the cylinder does not rotate, so that each point on it oves to the right with speed v CM. For the pure rotational otion shown in Figure.4b, iagine that a rotation ais through the center of ass is stationary, so that each point on the cylinder has the sae rotational speed. The cobination of these two otions represents the rolling otion shown in Figure.4c. Note in Figure.4c that the top of the cylinder has linear speed v CM R v CM v CM 2v CM, which is greater than the linear speed of any other point on the cylinder. As noted earlier, the center of ass oves with linear speed v CM while the contact point between the surface and cylinder has a linear speed of zero. We can epress the total kinetic energy of the rolling cylinder as K 2 I P2 (.3) where I P is the oent of inertia about a rotation ais through P. Applying the parallel-ais theore, we can substitute I P I CM MR 2 into Equation.3 to obtain K 2 I CM2 2MR 22 Q CM P P v CM 2v CM Figure.3 All points on a rolling object ove in a direction perpendicular to an ais through the instantaneous point of contact P. In other words, all points rotate about P. The center of ass of the object oves with a velocity v CM, and the point Poves with a velocity 2v CM. h Total kinetic energy of a rolling body M R Figure.5 A sphere rolling down an incline. Mechanical energy is conserved if no slipping occurs. θ ω v CM or, because v CM R, P K 2 I CM2 2 Mv 2 CM (.4) The ter 2 I CM2 represents the rotational kinetic energy of the cylinder about its center of ass, and the ter 2 Mv 2 CM represents the kinetic energy the cylinder would have if it were just translating through space without rotating. Thus, we can say that the total kinetic energy of a rolling object is the su of the rotational kinetic energy about the center of ass and the translational kinetic energy of the center of ass. We can use energy ethods to treat a class of probles concerning the rolling otion of a sphere down a rough incline. (The analysis that follows also applies to the rolling otion of a cylinder or hoop.) We assue that the sphere in Figure.5 rolls without slipping and is released fro rest at the top of the incline. Note that accelerated rolling otion is possible only if a frictional force is present between the sphere and the incline to produce a net torque about the center of ass. Despite the presence of friction, no loss of echanical energy occurs because the contact point is at rest relative to the surface at any instant. n the other hand, if the sphere were to slip, echanical energy would be lost as otion progressed. Using the fact that v CM R for pure rolling otion, we can epress Equation.4 as K 2 I CM v CM R 2 2 Mv 2 CM K 2 I CM R 2 v = 0 (c) Cobination of translation and rotation Figure.4 The otion of a rolling object can be odeled as a cobination of pure translation and pure rotation. M v 2 CM (.5)

44 By the tie the sphere reaches the botto of the incline, work equal to Mgh has been done on it by the gravitational field, where h is the height of the incline. Because the sphere starts fro rest at the top, its kinetic energy at the botto, given by Equation.5, ust equal this work done. Therefore, the speed of the center of ass at the botto can be obtained by equating these two quantities: Quick Quiz. 2 I CM R 2 M v 2 CM Mgh v CM 2gh 2/2 I CM /MR. Rolling Motion of a Rigid bject 33 (.6) Iagine that you slide your tetbook across a gynasiu floor with a certain initial speed. It quickly stops oving because of friction between it and the floor. Yet, if you were to start a basketball rolling with the sae initial speed, it would probably keep rolling fro one end of the gy to the other. Why does a basketball roll so far? Doesn t friction affect its otion? EXAMPLE. Sphere Rolling Down an Incline For the solid sphere shown in Figure.5, calculate the linear speed of the center of ass at the botto of the incline and the agnitude of the linear acceleration of the center of ass. The sphere starts fro the top of the incline with potential energy U g Mgh and kinetic energy K 0. As we have seen before, if it fell vertically fro that height, it would have a linear speed of!2gh at the oent before it hit the floor. After rolling down the incline, the linear speed of the center of ass ust be less than this value because soe of the initial potential energy is diverted into rotational kinetic energy rather than all being converted into translational kinetic energy. For a unifor solid sphere, I CM 2 5 MR 2 (see Table 0.2), and therefore Equation.6 gives 2gh /2 v 0 7 gh /2 CM 2/5MR 2 MR 2 which is less than!2gh. To calculate the linear acceleration of the center of ass, we note that the vertical displaceent is related to the displaceent along the incline through the relationship h EXAMPLE.2 Another Look at the Rolling Sphere In this eaple, let us use dynaic ethods to verify the results of Eaple.. The free-body diagra for the sphere is illustrated in Figure.6. ass gives Newton s second law applied to the center of sin. Hence, after squaring both sides, we can epress the equation above as v 2 CM 0 7 g sin Coparing this with the epression fro kineatics, v 2 CM 2a CM (see Eq. 2.2), we see that the acceleration of the center of ass is a CM 5 7 g sin These results are quite interesting in that both the speed and the acceleration of the center of ass are independent of the ass and the radius of the sphere! That is, all hoogeneous solid spheres eperience the sae speed and acceleration on a given incline. If we repeated the calculations for a hollow sphere, a solid cylinder, or a hoop, we would obtain siilar results in which only the factor in front of g sin would differ. The constant factors that appear in the epressions for v CM and a CM depend only on the oent of inertia about the center of ass for the specific body. In all cases, the acceleration of the center of ass is less than g sin, the value the acceleration would have if the incline were frictionless and no rolling occurred. () F Mg sin f Ma CM F y n Mg cos 0 where is easured along the slanted surface of the incline. Now let us write an epression for the torque acting on the sphere. A convenient ais to choose is the one that passes 332 CHAPTER Rolling Motion and Angular Moentu Figure.6 incline. through the center of the sphere and is perpendicular to the plane of the figure. Because n and Mg go through the center of ass, they have zero oent ar about this ais and thus do not contribute to the torque. However, the force of static friction produces a torque about this ais equal to fr in the clockwise direction; therefore, because is also in the QuickLab Hold a basketball and a tennis ball side by side at the top of a rap and release the at the sae tie. Which reaches the botto first? Does the outcoe depend on the angle of the rap? What if the angle were 90 (that is, if the balls were in free fall)? Torque y f Mg cos θ CM Mg n v CM Mg sin θ θ Free-body diagra for a solid sphere rolling down an 2.7 Quick Quiz.2 Which gets to the botto first: a ball rolling without sliding down incline A or a bo sliding down a frictionless incline B having the sae diensions as incline A?.2 clockwise direction, CM fr I CM Because I CM 2 5 MR 2 and we obtain (2) f I 2 CM R 5MR 2 R a CM R 2 5 Ma CM Substituting Equation (2) into Equation () gives a CM a CM /R, 5 7 g sin which agrees with the result of Eaple.. Note that F a applies only if F is the net eternal force eerted on the sphere and a is the acceleration of its center of ass. In the case of our sphere rolling down an incline, even though the frictional force does not change the total kinetic energy of the sphere, it does contribute to F and thus decreases the acceleration of the center of ass. As a result, the final translational kinetic energy is less than it would be in the absence of friction. As entioned in Eaple., soe of the initial potential energy is converted to rotational kinetic energy. THE VECTR PRDUCT AND TRQUE Consider a force F acting on a rigid body at the vector position r (Fig..7). The origin is assued to be in an inertial frae, so Newton s first law is valid in this case. As we saw in Section 0.6, the agnitude of the torque due to this force relative to the origin is, by definition, rf sin, where is the angle between r and F. The ais about which F tends to produce rotation is perpendicular to the plane fored by r and F. If the force lies in the y plane, as it does in Figure.7, the torque is represented by a vector parallel to the z ais. The force in Figure.7 creates a torque that tends to rotate the body counterclockwise about the z ais; thus the direction of is toward increasing z, and is therefore in the positive z direction. If we reversed the direction of F in Figure.7, then would be in the negative z direction. The torque involves the two vectors r and F, and its direction is perpendicular to the plane of r and F. We can establish a atheatical relationship between, r, and F, using a new atheatical operation called the vector product, or cross product: r F (.7) Although a coordinate syste whose origin is at the center of ass of a rolling object is not an inertial frae, the epression CM I still applies in the center-of-ass frae.

45 .2 The Vector Product and Torque CHAPTER Rolling Motion and Angular Moentu We now give a foral definition of the vector product. Given any two vectors A and B, the vector product A B is defined as a third vector C, the agnitude of which is AB sin, where is the angle between A and B. That is, if C is given by C A B (.8) then its agnitude is C AB sin (.9) The quantity AB sin is equal to the area of the parallelogra fored by A and B, as shown in Figure.8. The direction of C is perpendicular to the plane fored by A and B, and the best way to deterine this direction is to use the right-hand rule illustrated in Figure.8. The four fingers of the right hand are pointed along A and then wrapped into B through the angle. The direction of the erect right thub is the direction of A B C. Because of the notation, A B is often read A cross B ; hence, the ter cross product. Soe properties of the vector product that follow fro its definition are as follows:. Unlike the scalar product, the vector product is not coutative. Instead, the order in which the two vectors are ultiplied in a cross product is iportant: A B B A (.0) Therefore, if you change the order of the vectors in a cross product, you ust change the sign. You could easily verify this relationship with the right-hand rule. 2. If A is parallel to B ( 0 or 80 ), then A B 0; therefore, it follows that A A If A is perpendicular to B, then A B AB. 4. The vector product obeys the distributive law: A (B C) A B A C (.) 5. The derivative of the cross product with respect to soe variable such as t is d db (A B) A da B (.2) dt dt dt where it is iportant to preserve the ultiplicative order of A and B, in view of Equation.0. It is left as an eercise to show fro Equations.9 and.0 and fro the definition of unit vectors that the cross products of the rectangular unit vectors i, z τ = r F Figure.7 The torque vector lies in a direction perpendicular to the plane fored by the position vector r and the applied force vector F. Properties of the vector product r P φ y F Cross products of unit vectors EXAMPLE.3 j, and k obey the following rules: i i j j k k 0 (.3a) i j j i k (.3b) j k k j i (.3c) k i i k j (.3d) Signs are interchangeable in cross products. For eaple, A ( B) AB and i ( j) i j. The cross product of any two vectors A and B can be epressed in the following deterinant for: A B i j k A A y A z B B y B z i A y A z B y B z j A A z B B z k A A y B B y Epanding these deterinants gives the result The Cross Product Two vectors lying in the y plane are given by the equations A 2i 3j and B i 2j. Find A B and verify that A B B A. obtain Using Equations.3a through.3d, we A B (2i 3j) (i 2j) 2i 2j 3j (i) 4k 3k (We have oitted the ters containing i i and j j because, as Equation.3a shows, they are equal to zero.) We can show that A B B A, since A B (A y B z A z B y )i (A B z A z B )j (A B y A y B )k 7k B A (i 2j) (2i 3j) (.4) Therefore, A B B A. As an alternative ethod for finding A B, we could use Equation.4, with A 2, A y 3, A z 0 and B, B y 2, B z 0: A B (0)i (0)j [(2)(2) (3)()]k 7k Eercise Use the results to this eaple and Equation.9 to find the angle between A and B. Answer 60.3 i 3j 2j 2i 3k 4k 7k Right-hand rule.3 ANGULAR MMENTUM F A PARTICLE A θ B C = A B C = B A Figure.8 The vector product A B is a third vector C having a agnitude AB sin equal to the area of the parallelogra shown. The direction of C is perpendicular to the plane fored by A and B, and this direction is deterined by the righthand rule. 7.8 Iagine a rigid pole sticking up through the ice on a frozen pond (Fig..9). A skater glides rapidly toward the pole, aiing a little to the side so that she does not hit it. As she approaches a point beside the pole, she reaches out and grabs the pole, an action that whips her rapidly into a circular path around the pole. Just as the idea of linear oentu helps us analyze translational otion, a rotational analog angular oentu helps us describe this skater and other objects undergoing rotational otion. To analyze the otion of the skater, we need to know her ass and her velocity, as well as her position relative to the pole. In ore general ters, consider a

46 .3 Angular Moentu of a Particle CHAPTER Rolling Motion and Angular Moentu particle of ass located at the vector position r and oving with linear velocity v (Fig..0). The instantaneous angular oentu L of the particle relative to the origin is defined as the cross product of the particle s instantaneous position vector r and its instantaneous linear oentu p: L r p (.5) Angular oentu of a particle which is the rotational analog of Newton s second law, F dp/dt. Note that torque causes the angular oentu L to change just as force causes linear oentu p to change. This rotational result, Equation.9, states that the net torque acting on a particle is equal to the tie rate of change of the particle s angular oentu. The SI unit of angular oentu is kg 2 /s. It is iportant to note that both the agnitude and the direction of L depend on the choice of origin. Following the right-hand rule, note that the direction of L is perpendicular to the plane fored by r and p. In Figure.0, r and p are in the y plane, and so L points in the z direction. Because p v, the agnitude of L is L vr sin (.6) where is the angle between r and p. It follows that L is zero when r is parallel to p ( 0 or 80 ). In other words, when the linear velocity of the particle is along a line that passes through the origin, the particle has zero angular oentu with respect to the origin. n the other hand, if r is perpendicular to p ( 90 ), then L vr. At that instant, the particle oves eactly as if it were on the ri of a wheel rotating about the origin in a plane defined by r and p. Quick Quiz.3 Recall the skater described at the beginning of this section. What would be her angular oentu relative to the pole if she were skating directly toward it? In describing linear otion, we found that the net force on a particle equals the tie rate of change of its linear oentu, F dp/dt (see Eq. 9.3). We now show that the net torque acting on a particle equals the tie rate of change of its angular oentu. Let us start by writing the net torque on the particle in the for (.7) dt Now let us differentiate Equation.5 with respect to tie, using the rule given by Equation.2: dl dt d dt r F r dp (r p) r dp dt Reeber, it is iportant to adhere to the order of ters because A B B A. The last ter on the right in the above equation is zero because v dr/dt is parallel to p v (property 2 of the vector product). Therefore, d L dt r dp dt Coparing Equations.7 and.8, we see that d L dt dr p dt (.8) (.9) Figure.9 As the skater passes the pole, she grabs hold of it. This causes her to swing around the pole rapidly in a circular path. z L = r p r p Figure.0 The angular oentu L of a particle of ass and linear oentu p located at the vector position r is a vector given by L r p. The value of L depends on the origin about which it is easured and is a vector perpendicular to both r and p. The net torque equals tie rate of change of angular oentu φ y It is iportant to note that Equation.9 is valid only if and L are easured about the sae origin. (f course, the sae origin ust be used in calculating all of the torques.) Furtherore, the epression is valid for any origin fied in an inertial frae. Angular Moentu of a Syste of Particles The total angular oentu of a syste of particles about soe point is defined as the vector su of the angular oenta of the individual particles: where the vector su is over all n particles in the syste. Because individual angular oenta can change with tie, so can the total angular oentu. In fact, fro Equations.8 and.9, we find that the tie rate of change of the total angular oentu equals the vector su of all torques acting on the syste, both those associated with internal forces between particles and those associated with eternal forces. However, the net torque associated with all internal forces is zero. To understand this, recall that Newton s third law tells us that internal forces between particles of the syste are equal in agnitude and opposite in direction. If we assue that these forces lie along the line of separation of each pair of particles, then the torque due to each action reaction force pair is zero. That is, the oent ar d fro to the line of action of the forces is equal for both particles. In the suation, therefore, we see that the net internal torque vanishes. We conclude that the total angular oentu of a syste can vary with tie only if a net eternal torque is acting on the syste, so that we have That is, L L L 2 L n i L i et i dl i dt d dt i L i dl dt (.20) the tie rate of change of the total angular oentu of a syste about soe origin in an inertial frae equals the net eternal torque acting on the syste about that origin. Note that Equation.20 is the rotational analog of Equation 9.38, F et dp/dt, for a syste of particles.

47 .4 Angular Moentu of a Rotating Rigid bject CHAPTER Rolling Motion and Angular Moentu EXAMPLE.4 Circular Motion A particle oves in the y plane in a circular path of radius r, as shown in Figure.. (a) Find the agnitude and direction of its angular oentu relative to when its linear velocity is v. You ight guess that because the linear oentu of the particle is always changing (in direction, not agnitude), the direction of the angular oentu ust also change. In this eaple, however, this is not the case. The agnitude of L is given by L vr sin 90 vr (for r perpendicular to v) This value of L is constant because all three factors on the right are constant. The direction of L also is constant, even y v though the direction of p v keeps changing. You can visualize this by sliding the vector v in Figure. parallel to itself until its tail eets the tail of r and by then applying the right-hand rule. (You can use v to deterine the direction of L r p because the direction of p is the sae as the direction of v.) Line up your fingers so that they point along r and wrap your fingers into the vector v. Your thub points upward and away fro the page; this is the direction of L. Hence, we can write the vector epression L (vr)k. If the particle were to ove clockwise, L would point downward and into the page. (b) Find the agnitude and direction of L in ters of the particle s angular speed. Because v r for a particle rotating in a circle, we can epress L as L vr r 2 I z Figure.2 When a rigid body rotates about an ais, the angular oentu L is in the sae direction as the angular velocity, according to the epression L I. L r i ω v i y We can now find the angular oentu (which in this situation has only a z coponent) of the whole object by taking the su of L i over all particles: L z i i r i L z I (.2) where I is the oent of inertia of the object about the z ais. Now let us differentiate Equation.2 with respect to tie, noting that I is constant for a rigid body: dl z dt (.22) where is the angular acceleration relative to the ais of rotation. Because dl z /dt is equal to the net eternal torque (see Eq..20), we can epress Equation.22 as et dl z dt 2 i i r i I d dt I I 2 (.23) Figure. r A particle oving in a circle of radius r has an angular oentu about that has agnitude vr. The vector L r p points out of the diagra. where I is the oent of inertia of the particle about the z ais through. Because the rotation is counterclockwise, the direction of is along the z ais (see Section 0.). The direction of L is the sae as that of, and so we can write the angular oentu as L I Ik. Eercise A car of ass 500 kg oves with a linear speed of 40 /s on a circular race track of radius 50. What is the agnitude of its angular oentu relative to the center of the track? Answer kg 2 /s That is, the net eternal torque acting on a rigid object rotating about a fied ais equals the oent of inertia about the rotation ais ultiplied by the object s angular acceleration relative to that ais. Equation.23 also is valid for a rigid object rotating about a oving ais provided the oving ais () passes through the center of ass and (2) is a syetry ais. You should note that if a syetrical object rotates about a fied ais passing through its center of ass, you can write Equation.2 in vector for as L I, where L is the total angular oentu of the object easured with respect to the ais of rotation. Furtherore, the epression is valid for any object, regardless of its syetry, if L stands for the coponent of angular oentu along the ais of rotation. 2.4 ANGULAR MMENTUM F A RTATING RIGID BJECT Consider a rigid object rotating about a fied ais that coincides with the z ais of a coordinate syste, as shown in Figure.2. Let us deterine the angular oentu of this object. Each particle of the object rotates in the y plane about the z ais with an angular speed. The agnitude of the angular oentu of a particle of ass i about the origin is i v i r i. Because v i r i, we can epress the agnitude of the angular oentu of this particle as L i i r i The vector L i is directed along the z ais, as is the vector. 2 EXAMPLE.5 Bowling Ball Estiate the agnitude of the angular oentu of a bowling ball spinning at 0 rev/s, as shown in Figure.3. We start by aking soe estiates of the relevant physical paraeters and odel the ball as a unifor solid sphere. A typical bowling ball ight have a ass of 6 kg and a radius of 2 c. The oent of inertia of a solid sphere about an ais through its center is, fro Table 0.2, I 2 5 MR (6 kg)(0.2 ) kg 2 Therefore, the agnitude of the angular oentu is 2 In general, the epression L I is not always valid. If a rigid object rotates about an arbitrary ais, L and ay point in different directions. In this case, the oent of inertia cannot be treated as a scalar. Strictly speaking, L I applies only to rigid objects of any shape that rotate about one of three utually perpendicular aes (called principal aes) through the center of ass. This is discussed in ore advanced tets on echanics.

48 .4 Angular Moentu of a Rotating Rigid bject CHAPTER Rolling Motion and Angular Moentu Because of the roughness of our estiates, we probably want to keep only one significant figure, and so L 2 kg 2 /s. Figure.3 L I (0.035 kg 2 )(0 rev/s)(2 rad/rev) 2.2 kg 2 /s A bowling ball that rotates about the z ais in the direction shown has an angular oentu L in the positive z direction. If the direction of rotation is reversed, L points in the negative z direction. EXAMPLE.6 Rotating Rod A rigid rod of ass M and length is pivoted without friction at its center (Fig..4). Two particles of asses and 2 are connected to its ends. The cobination rotates in a vertical plane with an angular speed. (a) Find an epression for the agnitude of the angular oentu of the syste. This is different fro the last eaple in that we now ust account for the otion of ore than one object. The oent of inertia of the syste equals the su of the oents of inertia of the three coponents: the rod and the two particles. Referring to Table 0.2 to obtain the epression for the oent of inertia of the rod, and using the epression I r 2 for each particle, we find that the total oent of inertia about the z ais through is Therefore, the agnitude of the angular oentu is (b) Find an epression for the agnitude of the angular acceleration of the syste when the rod akes an angle with the horizontal. I 2 M M 3 2 L I If the asses of the two particles are equal, then the syste has no angular acceleration because the net torque on the syste is zero when 2. If the initial angle is eactly /2 or /2 (vertical position), then the rod will be in equilibriu. To find the angular acceleration of the syste at any angle, we first calculate the net torque on the syste and then use et I to obtain an epression for. The torque due to the force g about the pivot is g 2 cos 2 4 M 3 2 ( out of page) Figure.4 The torque due to the force 2 g about the pivot is z L y ( 2 into page) Hence, the net torque eerted on the syste about is The direction of et is out of the page if 2 and is into the page if 2. To find, we use et I, where I was obtained in part (a): Note that is zero when is /2 or /2 (vertical position) and is a aiu when is 0 or (horizontal position). Eercise If 2, at what value of is a aiu? Answer 2 2 g 2 cos et 2 2 ( 2 )g cos et I /2. g y θ 2 2 g Because gravitational forces act on the rotating rod, there is in general a net nonzero torque about when 2. This net torque produces an angular acceleration given by et I. 2( 2 )g cos (M/3 2 ) EXAMPLE.7 Two Connected Masses A sphere of ass and a block of ass 2 are connected by a light cord that passes over a pulley, as shown in Figure.5. The radius of the pulley is R, and the oent of inertia about its ale is I. The block slides on a frictionless, horizontal surface. Find an epression for the linear acceleration of the two objects, using the concepts of angular oentu and torque. We need to deterine the angular oentu of the syste, which consists of the two objects and the pulley. Let us calculate the angular oentu about an ais that coincides with the ale of the pulley. At the instant the sphere and block have a coon speed v, the angular oentu of the sphere is vr, and that of the block is 2 vr. At the sae instant, the angular oentu of the pulley is I Iv/R. Hence, the total angular oentu of the syste is () L vr 2 vr I v R R v Conservation of angular oentu v Figure Now let us evaluate the total eternal torque acting on the syste about the pulley ale. Because it has a oent ar of zero, the force eerted by the ale on the pulley does not contribute to the torque. Furtherore, the noral force acting on the block is balanced by the force of gravity 2 g, and so these forces do not contribute to the torque. The force of gravity g acting on the sphere produces a torque about the ale equal in agnitude to gr, where R is the oent ar of the force about the ale. (Note that in this situation, the tension is not equal to g.) This is the total eternal torque about the pulley ale; that is, et gr. Using this result, together with Equation () and Equation.23, we find This follows directly fro Equation.20, which indicates that if (2) et dl dt gr d dt ( 2 )Rv I v R gr ( 2 )R dv I dt R Because dv/dt a, we can solve this for a to obtain et dl 0 (.24) dt then L constant (.25) For a syste of particles, we write this conservation law as L n constant, where the inde n denotes the nth particle in the syste. a g ( 2 ) I/R 2 dv dt You ay wonder why we did not include the forces that the cord eerts on the objects in evaluating the net torque about the ale. The reason is that these forces are internal to the syste under consideration, and we analyzed the syste as a whole. nly eternal torques contribute to the change in the syste s angular oentu. CNSERVATIN F ANGULAR MMENTUM In Chapter 9 we found that the total linear oentu of a syste of particles reains constant when the resultant eternal force acting on the syste is zero. We have an analogous conservation law in rotational otion: The total angular oentu of a syste is constant in both agnitude and direction if the resultant eternal torque acting on the syste is zero.

49 .5 Conservation of Angular Moentu CHAPTER Rolling Motion and Angular Moentu If the ass of an object undergoes redistribution in soe way, then the object s oent of inertia changes; hence, its angular speed ust change because L I. In this case we epress the conservation of angular oentu in the for L i L f constant (.26) If the syste is an object rotating about a fied ais, such as the z ais, we can write L z I, where L z is the coponent of L along the ais of rotation and I is the oent of inertia about this ais. In this case, we can epress the conservation of angular oentu as I ii I f f constant (.27) This epression is valid both for rotation about a fied ais and for rotation about an ais through the center of ass of a oving syste as long as that ais reains parallel to itself. We require only that the net eternal torque be zero. Although we do not prove it here, there is an iportant theore concerning the angular oentu of an object relative to the object s center of ass: The resultant torque acting on an object about an ais through the center of ass equals the tie rate of change of angular oentu regardless of the otion of the center of ass. This theore applies even if the center of ass is accelerating, provided and L are evaluated relative to the center of ass. In Equation.26 we have a third conservation law to add to our list. We can now state that the energy, linear oentu, and angular oentu of an isolated syste all reain constant: For an isolated syste There are any eaples that deonstrate conservation of angular oentu. You ay have observed a figure skater spinning in the finale of a progra. The angular speed of the skater increases when the skater pulls his hands and feet close to his body, thereby decreasing I. Neglecting friction between skates and ice, no eternal torques act on the skater. The change in angular speed is due to the fact that, because angular oentu is conserved, the product I reains constant, and a decrease in the oent of inertia of the skater causes an increase in the angular speed. Siilarly, when divers or acrobats wish to ake several soersaults, they pull their hands and feet close to their bodies to rotate at a higher rate. In these cases, the eternal force due to gravity acts through the center of ass and hence eerts no torque about this point. Therefore, the angular oentu about the center of ass ust be conserved that is, I ii I f f. For eaple, when divers wish to double their angular speed, they ust reduce their oent of inertia to one-half its initial value. Quick Quiz.4 K i U i K f U f p i p f L i L f A particle oves in a straight line, and you are told that the net torque acting on it is zero about soe unspecified point. Decide whether the following stateents are true or false: (a) The net force on the particle ust be zero. (b) The particle s velocity ust be constant. Angular oentu is conserved as figure skater Todd Eldredge pulls his ars toward his body. ( 998 David Madison) EXAMPLE.8 Foration of a Neutron Star A star rotates with a period of 30 days about an ais through its center. After the star undergoes a supernova eplosion, the stellar core, which had a radius of k, collapses into a neutron star of radius 3.0 k. Deterine the period of rotation of the neutron star. The sae physics that akes a skater spin faster with his ars pulled in describes the otion of the neutron star. Let us assue that during the collapse of the stellar core, () no torque acts on it, (2) it reains spherical, and (3) its ass reains constant. Also, let us use the sybol T for the period, with T i being the initial period of the star and T f being the period of the neutron star. The period is the length EXAMPLE.9 The Merry-Go-Round A horizontal platfor in the shape of a circular disk rotates in a horizontal plane about a frictionless vertical ale (Fig..6). The platfor has a ass M 00 kg and a radius R 2.0. A student whose ass is 60 kg walks slowly fro the ri of the disk toward its center. If the angular speed of the syste is 2.0 rad/s when the student is at the ri, what is the angular speed when he has reached a point r 0.50 fro the center? A color-enhanced, infrared iage of Hurricane Mitch, which devastated large areas of Honduras and Nicaragua in ctober 998. The spiral, nonrigid ass of air undergoes rotation and has angular oentu. (Courtesy of NAA) of tie a point on the star s equator takes to ake one coplete circle around the ais of rotation. The angular speed of a star is given by 2/T. Therefore, because I is proportional to r 2, Equation.27 gives T f T i r f r i 2 (30 days) 3.0 k k days 0.23 s Thus, the neutron star rotates about four ties each second; this result is approiately the sae as that for a spinning figure skater. The speed change here is siilar to the increase in angular speed of the spinning skater when he pulls his ars inward. Let us denote the oent of inertia of the platfor as I p and that of the student as I s. Treating the student as a point ass, we can write the initial oent of inertia I i of the syste (student plus platfor) about the ais of rotation: I i I pi I si 2 MR 2 R 2

50 .5 Conservation of Angular Moentu CHAPTER Rolling Motion and Angular Moentu Figure.6 M R As the student walks toward the center of the rotating platfor, the angular speed of the syste increases because the angular oentu ust reain constant. Quick Quiz.5 EXAMPLE.0 The Spinning Bicycle Wheel In a favorite classroo deonstration, a student holds the ale of a spinning bicycle wheel while seated on a stool that is free to rotate (Fig..7). The student and stool are initially at rest while the wheel is spinning in a horizontal plane with an initial angular oentu L i that points upward. When the wheel is inverted about its center by 80, the student and Figure.7 The wheel is initially spinning when the student is at rest. What happens when the wheel is inverted? L i When the student has walked to the position r R, the oent of inertia of the syste reduces to I f I pf I sf 2 MR 2 r 2 Note that we still use the greater radius R when calculating I pf because the radius of the platfor has not changed. Because no eternal torques act on the syste about the ais of rotation, we can apply the law of conservation of angular oentu: I ii I f f 2 MR 2 R 2 i ( 2 MR 2 r 2 )f f 2 MR 2 R 2 2 MR 2 r 2 i f (2.0 rad/s) As epected, the angular speed has increased. Note that the rotational energy of the syste described in Eaple.9 increases. What accounts for this increase in energy? 4. rad/s Eercise Calculate the initial and final rotational energies of the syste. Answer K i 880 J; K f J. stool start rotating. In ters of L i, what are the agnitude and the direction of L for the student plus stool? The syste consists of the student, the wheel, and the stool. Initially, the total angular oentu of the syste L i coes entirely fro the spinning wheel. As the wheel is inverted, the student applies a torque to the wheel, but this torque is internal to the syste. No eternal torque is acting on the syste about the vertical ais. Therefore, the angular oentu of the syste is conserved. Initially, we have L syste L i L wheel (upward) After the wheel is inverted, we have L inverted wheel L i. For angular oentu to be conserved, soe other part of the syste has to start rotating so that the total angular oentu reains the initial angular oentu L i. That other part of the syste is the student plus the stool she is sitting on. So, we can now state that L f L i L studentstool L i L studentstool 2L i EXAMPLE. Disk and Stick A 2.0-kg disk traveling at 3.0 /s strikes a.0-kg stick that is lying flat on nearly frictionless ice, as shown in Figure.8. Assue that the collision is elastic. Find the translational speed of the disk, the translational speed of the stick, and the rotational speed of the stick after the collision. The oent of inertia of the stick about its center of ass is.33 kg 2. Because the disk and stick for an isolated syste, we can assue that total energy, linear oentu, and angular oentu are all conserved. We have three unknowns, and so we need three equations to solve siultaneously. The first coes fro the law of the conservation of linear oentu: p i p f d v di d v d f s v s (2.0 kg)(3.0 /s) (2.0 kg)v d f (.0 kg)v s () 6.0 kg/s (2.0 kg)v d f (.0 kg)v s Now we apply the law of conservation of angular oentu, using the initial position of the center of the stick as our reference point. We know that the coponent of angular oentu of the disk along the ais perpendicular to the plane of the ice is negative (the right-hand rule shows that L d points into the ice). L i L f r d v di r d v d f I (2.0 )(2.0 kg)(3.0 /s) (2.0 )(2.0 kg)v d f (.33 kg 2 ) 2 kg 2 /s (4.0 kg)v d f (.33 kg 2 ) (2) 9.0 rad/s (3.0 rad/)v d f TABLE. We used the fact that radians are diensionless to ensure consistent units for each ter. Finally, the elastic nature of the collision reinds us that kinetic energy is conserved; in this case, the kinetic energy consists of translational and rotational fors: K i K f 2 dv 2 di 2 dv 2 d f 2 sv 2 s 2 I2 2 (2.0 kg)(3.0 /s)2 2 (2.0 kg)v d f 2 2 (.0 kg)v 2 s 2 (.33 kg2 /s)2 (3) 54 2 /s 2 6.0v 2 d f 3.0v 2 s (4.0 2 )2 In solving Equations (), (2), and (3) siultaneously, we find that v df 2.3 /s, v s.3 /s, and 2.0 rad/s. These values see reasonable. The disk is oving ore slowly than it was before the collision, and the stick has a sall translational speed. Table. suarizes the initial and final values of variables for the disk and the stick and verifies the conservation of linear oentu, angular oentu, and kinetic energy. Eercise Verify the values in Table Before Figure.8 v di = 3.0 /s Coparison of Values in Eaple. Before and After the Collision a Ktrans Krot v (/s) (rad/s) p (kg/s) L (kg 2 /s) ( J) ( J) Before Disk Stick Total After Disk Stick Total After verhead view of a disk striking a stick in an elastic collision, which causes the stick to rotate. a Notice that linear oentu, angular oentu, and total kinetic energy are conserved. v s v df ω

51 .6 The Motion of Gyroscopes and Tops CHAPTER Rolling Motion and Angular Moentu ptional Section.6 THE MTIN F GYRSCPES AND TPS A very unusual and fascinating type of otion you probably have observed is that of a top spinning about its ais of syetry, as shown in Figure.9a. If the top spins very rapidly, the ais rotates about the z ais, sweeping out a cone (see Fig..9b). The otion of the ais about the vertical known as precessional otion is usually slow relative to the spin otion of the top. It is quite natural to wonder why the top does not fall over. Because the center of ass is not directly above the pivot point, a net torque is clearly acting on the top about a torque resulting fro the force of gravity Mg. The top would certainly fall over if it were not spinning. Because it is spinning, however, it has an angular oentu L directed along its syetry ais. As we shall show, the otion of this syetry ais about the z ais (the precessional otion) occurs because the torque produces a change in the direction of the syetry ais. This is an ecellent eaple of the iportance of the directional nature of angular oentu. The two forces acting on the top are the downward force of gravity Mg and the noral force n acting upward at the pivot point. The noral force produces no torque about the pivot because its oent ar through that point is zero. However, the force of gravity produces a torque r Mg about, where the direction of is perpendicular to the plane fored by r and Mg. By necessity, the vector lies in a horizontal y plane perpendicular to the angular oentu vector. The net torque and angular oentu of the top are related through Equation.9: dl Fro this epression, we see that the nonzero torque produces a change in angular oentu dl a change that is in the sae direction as. Therefore, like the torque vector, dl ust also be at right angles to L. Figure.9b illustrates the resulting precessional otion of the syetry ais of the top. In a tie t, the change in angular oentu is L L f L i Because L is perpendicular to L, the agnitude of L does not change ( L i L f ). Rather, what is changing is the direction of L. Because the change in angular oentu L is in the direction of, which lies in the y plane, the top undergoes precessional otion. The essential features of precessional otion can be illustrated by considering the siple gyroscope shown in Figure.20a. This device consists of a wheel free to spin about an ale that is pivoted at a distance h fro the center of ass of the wheel. When given an angular velocity about the ale, the wheel has an angular oentu L I directed along the ale as shown. Let us consider the torque acting on the wheel about the pivot. Again, the force n eerted by the support on the ale produces no torque about, and the force of gravity Mg produces a torque of agnitude Mgh about, where the ale is perpendicular to the support. The direction of this torque is perpendicular to the ale (and perpendicular to L), as shown in Figure.20a. This torque causes the angular oentu to change in the direction perpendicular to the ale. Hence, the ale oves in the direction of the torque that is, in the horizontal plane. To siplify the description of the syste, we ust ake an assuption: The total angular oentu of the precessing wheel is the su of the angular oentu I due to the spinning and the angular oentu due to the otion of dt t. Precessional otion L Mg CM L i r L z L f n τ (a) y (b) Figure.9 Precessional otion of a top spinning about its syetry ais. (a) The only eternal forces acting on the top are the noral force n and the force of gravity Mg. The direction of the angular oentu L is along the ais of syetry. The right-hand rule indicates that r F r Mg is in the y plane. (b). The direction of L is parallel to that of in part (a). The fact that L f L L i indicates that the top precesses about the z ais. L i L f Mg (a) h n Figure.20 (a) The otion of a siple gyroscope pivoted a distance h fro its center of ass. The force of gravity Mg produces a torque about the pivot, and this torque is perpendicular to the ale. (b) This torque results in a change in angular oentu dl in a direction perpendicular to the ale. The ale sweeps out an angle d in a tie dt. n τ τ This toy gyroscope undergoes precessional otion about the vertical ais as it spins about its ais of syetry. The only forces acting on it are the force of gravity Mg and the upward force of the pivot n. The direction of its angular oentu L is along the ais of syetry. The torque and L are directed into the page. (Courtesy of Central Scientific Copany) the center of ass about the pivot. In our treatent, we shall neglect the contribution fro the center-of-ass otion and take the total angular oentu to be just I. In practice, this is a good approiation if is ade very large. In a tie dt, the torque due to the gravitational force changes the angular oentu of the syste by dl dt. When added vectorially to the original total r Mg d L L L i (b) dφ L f

52 .7 Angular Moentu as a Fundaental Quantity CHAPTER Rolling Motion and Angular Moentu angular oentu I, this additional angular oentu causes a shift in the direction of the total angular oentu. The vector diagra in Figure.20b shows that in the tie dt, the angular oentu vector rotates through an angle d, which is also the angle through which the ale rotates. Fro the vector triangle fored by the vectors L i, L f, and dl, we see that sin (d) d dl (Mgh)dt L L where we have used the fact that, for sall values of any angle, sin. Dividing through by dt and using the relationship L I, we find that the rate at which the ale rotates about the vertical ais is p d Mgh (.28) dt The angular speed p is called the precessional frequency. This result is valid only when p V. therwise, a uch ore coplicated otion is involved. As you can see fro Equation.28, the condition p V is et when I is great copared with Mgh. Furtherore, note that the precessional frequency decreases as increases that is, as the wheel spins faster about its ais of syetry. Quick Quiz.6 How uch work is done by the force of gravity when a top precesses through one coplete circle? ptional Section.7 ANGULAR MMENTUM AS A FUNDAMENTAL QUANTITY We have seen that the concept of angular oentu is very useful for describing the otion of acroscopic systes. However, the concept also is valid on a subicroscopic scale and has been used etensively in the developent of odern theories of atoic, olecular, and nuclear physics. In these developents, it was found that the angular oentu of a syste is a fundaental quantity. The word fundaental in this contet iplies that angular oentu is an intrinsic property of atos, olecules, and their constituents, a property that is a part of their very nature. To eplain the results of a variety of eperients on atoic and olecular systes, we rely on the fact that the angular oentu has discrete values. These discrete values are ultiples of the fundaental unit of angular oentu h/2, where h is called Planck s constant: Fundaental unit of angular oentu kg 2 /s Let us accept this postulate without proof for the tie being and show how it can be used to estiate the angular speed of a diatoic olecule. Consider the 2 olecule as a rigid rotor, that is, two atos separated by a fied distance d and rotating about the center of ass (Fig..2). Equating the angular oentu to the fundaental unit, we can estiate the lowest angular speed: I CM or I I CM Precessional frequency d CM Figure.2 The rigid-rotor odel of a diatoic olecule. The rotation occurs about the center of ass in the plane of the page. ω In Eaple 0.3, we found that the oent of inertia of the 2 olecule about this ais of rotation is kg 2. Therefore, kg 2 /s I CM kg rad/s Actual angular speeds are ultiples of this sallest possible value. This siple eaple shows that certain classical concepts and odels, when properly odified, ight be useful in describing soe features of atoic and olecular systes. A wide variety of phenoena on the subicroscopic scale can be eplained only if we assue discrete values of the angular oentu associated with a particular type of otion. The Danish physicist Niels Bohr ( ) accepted and adopted this radical idea of discrete angular oentu values in developing his theory of the hydrogen ato. Strictly classical odels were unsuccessful in describing any properties of the hydrogen ato. Bohr postulated that the electron could occupy only those circular orbits about the proton for which the orbital angular oentu was equal to n, where n is an integer. That is, he ade the bold assuption that orbital angular oentu is quantized. Fro this siple odel, the rotational frequencies of the electron in the various orbits can be estiated (see Proble 43). SUMMARY The total kinetic energy of a rigid object rolling on a rough surface without slipping equals the rotational kinetic energy about its center of ass, plus the translational kinetic energy of the center of ass, 2 Mv CM 2 2 I CM2, : K 2 I CM2 2 Mv 2 CM (.4) The torque due to a force F about an origin in an inertial frae is defined to be (.7) Given two vectors A and B, the cross product A B is a vector C having a agnitude C AB sin (.9) where is the angle between A and B. The direction of the vector C A B is perpendicular to the plane fored by A and B, and this direction is deterined by the right-hand rule. The angular oentu L of a particle having linear oentu p v is L r p (.5) where r is the vector position of the particle relative to an origin in an inertial frae. The net eternal torque acting on a particle or rigid object is equal to the tie rate of change of its angular oentu: et dl (.20) dt The z coponent of angular oentu of a rigid object rotating about a fied z ais is L z I (.2) r F

53 Questions CHAPTER Rolling Motion and Angular Moentu where I is the oent of inertia of the object about the ais of rotation and is its angular speed. The net eternal torque acting on a rigid object equals the product of its oent of inertia about the ais of rotation and its angular acceleration: et I (.23) If the net eternal torque acting on a syste is zero, then the total angular oentu of the syste is constant. Applying this law of conservation of angular oentu to a syste whose oent of inertia changes gives I ii I f f constant (.27) 7. Two solid spheres a large, assive sphere and a sall sphere with low ass are rolled down a hill. Which sphere reaches the botto of the hill first? Net, a large, low-density sphere and a sall, high-density sphere having the sae ass are rolled down the hill. Which one reaches the botto first in this case? 8. Suppose you are designing a car for a coasting race the cars in this race have no engines; they siply coast down a hill. Do you want to use large wheels or sall wheels? Do you want to use solid, disk-like wheels or hoop-like wheels? Should the wheels be heavy or light? 9. Why do tightrope walkers carry a long pole to help theselves keep their balance? 20. Two balls have the sae size and ass. ne is hollow, whereas the other is solid. How would you deterine which is which without breaking the apart? 2. A particle is oving in a circle with constant speed. Locate one point about which the particle s angular oentu is constant and another about which it changes with tie. 22. If global waring occurs over the net century, it is likely that soe polar ice will elt and the water will be distributed closer to the equator. How would this change the oent of inertia of the Earth? Would the length of the day (one revolution) increase or decrease? QUESTINS. Is it possible to calculate the torque acting on a rigid body without specifying a center of rotation? Is the torque independent of the location of the center of rotation? 2. Is the triple product defined by A (B C) a scalar or a vector quantity? Eplain why the operation (A B) C has no eaning. 3. In soe otorcycle races, the riders drive over sall hills, and the otorcycles becoe airborne for a short tie. If a otorcycle racer keeps the throttle open while leaving the hill and going into the air, the otorcycle tends to nose upward. Why does this happen? 4. If the torque acting on a particle about a certain origin is zero, what can you say about its angular oentu about that origin? 5. Suppose that the velocity vector of a particle is copletely specified. What can you conclude about the direction of its angular oentu vector with respect to the direction of otion? 6. If a single force acts on an object, and the torque caused by that force is nonzero about soe point, is there any other point about which the torque is zero? 7. If a syste of particles is in otion, is it possible for the total angular oentu to be zero about soe origin? Eplain. 8. A ball is thrown in such a way that it does not spin about its own ais. Does this ean that the angular oentu is zero about an arbitrary origin? Eplain. 9. In a tape recorder, the tape is pulled past the read-andwrite heads at a constant speed by the drive echanis. Consider the reel fro which the tape is pulled as the tape is pulled off it, the radius of the roll of reaining tape decreases. How does the torque on the reel change with tie? How does the angular speed of the reel change with tie? If the tape echanis is suddenly turned on so that the tape is quickly pulled with a great force, is the tape ore likely to break when pulled fro a nearly full reel or a nearly epty reel? 0. A scientist at a hotel sought assistance fro a bellhop to carry a ysterious suitcase. When the unaware bellhop rounded a corner carrying the suitcase, it suddenly oved away fro hi for soe unknown reason. At this point, the alared bellhop dropped the suitcase and ran off. What do you suppose ight have been in the suitcase?. When a cylinder rolls on a horizontal surface as in Figure.3, do any points on the cylinder have only a vertical coponent of velocity at soe instant? If so, where are they? 2. Three objects of unifor density a solid sphere, a solid cylinder, and a hollow cylinder are placed at the top of an incline (Fig. Q.2). If they all are released fro rest at the sae elevation and roll without slipping, which object reaches the botto first? Which reaches it last? You should try this at hoe and note that the result is independent of the asses and the radii of the objects. Figure Q.2 Which object wins the race? 3. A ouse is initially at rest on a horizontal turntable ounted on a frictionless vertical ale. If the ouse begins to walk around the perieter, what happens to the turntable? Eplain. 4. Stars originate as large bodies of slowly rotating gas. Because of gravity, these regions of gas slowly decrease in size. What happens to the angular speed of a star as it shrinks? Eplain. 5. ften, when a high diver wants to eecute a flip in idair, she draws her legs up against her chest. Why does this ake her rotate faster? What should she do when she wants to coe out of her flip? 6. As a tether ball winds around a thin pole, what happens to its angular speed? Eplain. PRBLEMS, 2, 3 = straightforward, interediate, challenging = full solution available in the Student s Manual and Study Guide WEB = solution posted at = Coputer useful in solving proble = Interactive Physics = paired nuerical/sybolic probles Section. Rolling Motion of a Rigid bject. A cylinder of ass 0.0 kg rolls without slipping on a horizontal surface. At the instant its center of ass has a speed of 0.0 /s, deterine (a) the translational kinetic energy of its center of ass, (b) the rotational energy about its center of ass, and (c) its total energy. 2. A bowling ball has a ass of 4.00 kg, a oent of inertia of kg 2, and a radius of If it rolls down the lane without slipping at a linear speed of 4.00 /s, what is its total energy? 3. A bowling ball has a ass M, a radius R, and a oent 2 of inertia 5 MR 2. If it starts fro rest, how uch work ust be done on it to set it rolling without slipping at a linear speed v? Epress the work in ters of M and v. 4. A unifor solid disk and a unifor hoop are placed side by side at the top of an incline of height h. If they are released fro rest and roll without slipping, deterine their speeds when they reach the botto. Which object reaches the botto first? 5. (a) Deterine the acceleration of the center of ass of a unifor solid disk rolling down an incline aking an angle with the horizontal. Copare this acceleration with that of a unifor hoop. (b) What is the iniu coefficient of friction required to aintain pure rolling otion for the disk? 6. A ring of ass 2.40 kg, inner radius 6.00 c, and outer radius 8.00 c rolls (without slipping) up an inclined plane that akes an angle of 36.9 (Fig. P.6). At the oent the ring is at position 2.00 up the plane, its speed is 2.80 /s. The ring continues up the plane for soe additional distance and then rolls back down. It does not roll off the top end. How far up the plane does it go? WEB θ Figure P.6 7. A etal can containing condensed ushroo soup has a ass of 25 g, a height of 0.8 c, and a diaeter of 6.38 c. It is placed at rest on its side at the top of a long incline that is at an angle of 25.0 to the horizontal and is then released to roll straight down. Assuing energy conservation, calculate the oent of inertia of the can if it takes.50 s to reach the botto of the incline. Which pieces of data, if any, are unnecessary for calculating the solution? 8. A tennis ball is a hollow sphere with a thin wall. It is set rolling without slipping at 4.03 /s on the horizontal section of a track, as shown in Figure P.8. It rolls around the inside of a vertical circular loop 90.0 c in diaeter and finally leaves the track at a point 20.0 c below the horizontal section. (a) Find the speed of the ball at the top of the loop. Deonstrate that it will not fall fro the track. (b) Find its speed as it leaves the track. (c) Suppose that static friction between the ball and the track was negligible, so that the ball slid instead of rolling. Would its speed v

54 Probles CHAPTER Rolling Motion and Angular Moentu B tu of the ass about the center of the circle is L ( 2 g 3 sin 4 /cos ) /2 v = v i i F 3 D v i θ v 2 R C A F 2 F θ Figure P.25 WEB Figure P.8 then be higher, lower, or the sae at the top of the loop? Eplain. Section.2 The Vector Product and Torque 9. Given M 6i 2j k and N 2i j 3k, calculate the vector product M N. 0. The vectors 42.0 c at 5.0 and 23.0 c at 65.0 both start fro the origin. Both angles are easured counterclockwise fro the ais. The vectors for two sides of a parallelogra. (a) Find the area of the parallelogra. (b) Find the length of its longer diagonal.. Two vectors are given by A 3i 4j and B 2i 3j. Find (a) A B and (b) the angle between A and B. 2. For the vectors A 3i 7j 4k and B 6i 0j 9k, evaluate the epressions (a) cos (A B/AB ) and (b) sin ( A B /AB). (c) Which give(s) the angle between the vectors? 3. A force of F 2.00i 3.00j N is applied to an object that is pivoted about a fied ais aligned along the z coordinate ais. If the force is applied at the point r (4.00i 5.00j 0k), find (a) the agnitude of the net torque about the z ais and (b) the direction of the torque vector. 4. A student clais that she has found a vector A such that (2i 3j 4k) A (4i 3j k). Do you believe this clai? Eplain. 5. Vector A is in the negative y direction, and vector B is in the negative direction. What are the directions of (a) A B and (b) B A? 6. A particle is located at the vector position r (i 3j), and the force acting on it is F (3i 2j) N. What is the torque about (a) the origin and (b) the point having coordinates (0, 6)? 7. If A B A B, what is the angle between A and B? 8. Two forces F and F 2 act along the two sides of an equilateral triangle, as shown in Figure P.8. Point is the intersection of the altitudes of the triangle. Find a third force F 3 to be applied at B and along BC that will ake the total torque about the point be zero. Will the total torque change if F 3 is applied not at B, but rather at any other point along BC? Section.3 Figure P.8 Angular Moentu of a Particle 9. A light, rigid rod.00 in length joins two particles with asses 4.00 kg and 3.00 kg at its ends. The cobination rotates in the y plane about a pivot through the center of the rod (Fig. P.9). Deterine the angular oentu of the syste about the origin when the speed of each particle is 5.00 /s kg v y v.00 Figure P kg 20. A.50-kg particle oves in the y plane with a velocity of v (4.20i 3.60j) /s. Deterine the particle s angular oentu when its position vector is r (.50i 2.20j). WEB 2. The position vector of a particle of ass 2.00 kg is given as a function of tie by r (6.00i 5.00t j). Deterine the angular oentu of the particle about the origin as a function of tie. 22. A conical pendulu consists of a bob of ass in otion in a circular path in a horizontal plane, as shown in Figure P.22. During the otion, the supporting wire of length aintains the constant angle with the vertical. Show that the agnitude of the angular oen- Figure P A particle of ass oves in a circle of radius R at a constant speed v, as shown in Figure P.23. If the otion begins at point Q, deterine the angular oentu of the particle about point P as a function of tie. P y Figure P A 4.00-kg ass is attached to a light cord that is wound around a pulley (see Fig. 0.20). The pulley is a unifor solid cylinder with a radius of 8.00 c and a ass of 2.00 kg. (a) What is the net torque on the syste about the point? (b) When the ass has a speed v, the pulley has an angular speed v/r. Deterine the total angular oentu of the syste about. (c) Using the fact that dl/dt and your result fro part (b), calculate the acceleration of the ass. 25. A particle of ass is shot with an initial velocity v i and akes an angle with the horizontal, as shown in Figure P.25. The particle oves in the gravitational field of the Earth. Find the angular oentu of the parti- R v Q cle about the origin when the particle is (a) at the origin, (b) at the highest point of its trajectory, and (c) just about to hit the ground. (d) What torque causes its angular oentu to change? 26. Heading straight toward the suit of Pike s Peak, an airplane of ass kg flies over the plains of Kansas at a nearly constant altitude of 4.30 k and with a constant velocity of 75 /s westward. (a) What is the airplane s vector angular oentu relative to a wheat farer on the ground directly below the airplane? (b) Does this value change as the airplane continues its otion along a straight line? (c) What is its angular oentu relative to the suit of Pike s Peak? 27. A ball of ass is fastened at the end of a flagpole connected to the side of a tall building at point P, as shown in Figure P.27. The length of the flagpole is, and is the angle the flagpole akes with the horizontal. Suppose that the ball becoes loose and starts to fall. Deterine the angular oentu (as a function of tie) of the ball about point P. Neglect air resistance. P Figure P A firean clings to a vertical ladder and directs the nozzle of a hose horizontally toward a burning building. The rate of water flow is 6.3 kg/s, and the nozzle speed is 2.5 /s. The hose passes between the firean s feet, which are.30 vertically below the nozzle. Choose the origin to be inside the hose between the firean s θ

55 Probles CHAPTER Rolling Motion and Angular Moentu feet. What torque ust the firean eert on the hose? That is, what is the rate of change of angular oentu of the water? Section.4 Angular Moentu of a Rotating Rigid bject 29. A unifor solid sphere with a radius of and a ass of 5.0 kg turns counterclockwise about a vertical ais through its center. Find its vector angular oentu when its angular speed is 3.00 rad/s. 30. A unifor solid disk with a ass of 3.00 kg and a radius of rotates about a fied ais perpendicular to its face. If the angular speed is 6.00 rad/s, calculate the angular oentu of the disk when the ais of rotation (a) passes through its center of ass and (b) passes through a point idway between the center and the ri. 3. A particle with a ass of kg is attached to the 00-c ark of a eter stick with a ass of 0.00 kg. The eter stick rotates on a horizontal, frictionless table with an angular speed of 4.00 rad/s. Calculate the angular oentu of the syste when the stick is pivoted about an ais (a) perpendicular to the table through the 50.0-c ark and (b) perpendicular to the table through the 0-c ark. 32. The hour and inute hands of Big Ben, the faous Parliaent Building tower clock in London, are 2.70 and 4.50 long and have asses of 60.0 kg and 00 kg, respectively. Calculate the total angular oentu of these hands about the center point. Treat the hands as long thin rods. Section.5 Conservation of Angular Moentu 33. A cylinder with a oent of inertia of I rotates about a vertical, frictionless ale with angular velocity i. A second cylinder that has a oent of inertia of I 2 and initially is not rotating drops onto the first cylinder (Fig. P.33). Because of friction between the surfaces, the two eventually reach the sae angular speed f. (a) Calculate f. (b) Show that the kinetic energy of the syste decreases in this interaction and calculate I I 2 Before ω i Figure P.33 After ω f WEB the ratio of the final rotational energy to the initial rotational energy. 34. A playground erry-go-round of radius R 2.00 has a oent of inertia of I 250 kg 2 and is rotating at 0.0 rev/in about a frictionless vertical ale. Facing the ale, a 25.0-kg child hops onto the erry-go-round and anages to sit down on its edge. What is the new angular speed of the erry-go-round? 35. A student sits on a freely rotating stool holding two weights, each of which has a ass of 3.00 kg. When his ars are etended horizontally, the weights are.00 fro the ais of rotation and he rotates with an angular speed of rad/s. The oent of inertia of the student plus stool is 3.00 kg 2 and is assued to be constant. The student pulls the weights inward horizontally to a position fro the rotation ais. (a) Find the new angular speed of the student. (b) Find the kinetic energy of the rotating syste before and after he pulls the weights inward. 36. A unifor rod with a ass of 00 g and a length of 50.0 c rotates in a horizontal plane about a fied, vertical, frictionless pin passing through its center. Two sall beads, each having a ass 30.0 g, are ounted on the rod so that they are able to slide without friction along its length. Initially, the beads are held by catches at positions 0.0 c on each side of center; at this tie, the syste rotates at an angular speed of 20.0 rad/s. Suddenly, the catches are released, and the sall beads slide outward along the rod. Find (a) the angular speed of the syste at the instant the beads reach the ends of the rod and (b) the angular speed of the rod after the beads fly off the rod s ends. 37. A 60.0-kg woan stands at the ri of a horizontal turntable having a oent of inertia of 500 kg 2 and a radius of The turntable is initially at rest and is free to rotate about a frictionless, vertical ale through its center. The woan then starts walking around the ri clockwise (as viewed fro above the syste) at a constant speed of.50 /s relative to the Earth. (a) In what direction and with what angular speed does the turntable rotate? (b) How uch work does the woan do to set herself and the turntable into otion? 38. A puck with a ass of 80.0 g and a radius of 4.00 c slides along an air table at a speed of.50 /s, as shown in Figure P.38a. It akes a glancing collision.50 /s (a) Figure P.38 (b) with a second puck having a radius of 6.00 c and a ass of 20 g (initially at rest) such that their ris just touch. Because their ris are coated with instant-acting glue, the pucks stick together and spin after the collision (Fig. P.38b). (a) What is the angular oentu of the syste relative to the center of ass? (b) What is the angular speed about the center of ass? 39. A wooden block of ass M resting on a frictionless horizontal surface is attached to a rigid rod of length and of negligible ass (Fig. P.39). The rod is pivoted at the other end. A bullet of ass traveling parallel to the horizontal surface and noral to the rod with speed v hits the block and becoes ebedded in it. (a) What is the angular oentu of the bullet block syste? (b) What fraction of the original kinetic energy is lost in the collision? M 40. A space station shaped like a giant wheel has a radius of 00 and a oent of inertia of kg 2. A crew of 50 are living on the ri, and the station s rotation causes the crew to eperience an acceleration of g (Fig. P.40). When 00 people ove to the center of the station for a union eeting, the angular speed changes. What acceleration is eperienced by the anagers reaining at the ri? Assue that the average ass of each inhabitant is 65.0 kg. 4. A wad of sticky clay of ass and velocity v i is fired at a solid cylinder of ass M and radius R (Fig. P.4). The cylinder is initially at rest and is ounted on a fied horizontal ale that runs through the center of ass. The line of otion of the projectile is perpendicular to the ale and at a distance d, less than R, fro the center. (a) Find the angular speed of the syste just after the clay strikes and sticks to the surface of the cylinder. (b) Is echanical energy conserved in this process? Eplain your answer. 42. Suppose a eteor with a ass of kg is oving at 30.0 k/s relative to the center of the Earth and strikes the Earth. What is the order of agnitude of the v Figure P.39 d v i Figure P.40 aiu possible decrease in the angular speed of the Earth due to this collision? Eplain your answer. (ptional) Section.7 Angular Moentu as a Fundaental Quantity 43. In the Bohr odel of the hydrogen ato, the electron oves in a circular orbit of radius around the proton. Assuing that the orbital angular oentu of the electron is equal to h/2, calculate (a) the orbital speed of the electron, (b) the kinetic energy of the electron, and (c) the angular speed of the electron s otion. ADDITINAL PRBLEMS 44. Review Proble. A rigid, assless rod has three equal asses attached to it, as shown in Figure P.44. The rod is free to rotate in a vertical plane about a frictionless ale perpendicular to the rod through the point P, and it is released fro rest in the horizontal position at t 0. Assuing and d are known, find (a) the oent of inertia of the syste about the pivot, (b) the torque acting on the syste at t 0, (c) the angular acceleration of the syste at t 0, (d) the linear acceleration of the ass labeled 3 at t 0, (e) the aiu R Figure P.4 M

56 Probles CHAPTER Rolling Motion and Angular Moentu kinetic energy of the syste, (f) the aiu angular speed attained by the rod, (g) the aiu angular oentu of the syste, and (h) the aiu speed attained by the ass labeled d d 45. A unifor solid sphere of radius r is placed on the inside surface of a heispherical bowl having a uch greater radius R. The sphere is released fro rest at an angle to the vertical and rolls without slipping (Fig. P.45). Deterine the angular speed of the sphere when it reaches the botto of the bowl. P Figure P.44 2d 3 tie. (f) Find the work done by the drive otor during the 440-s otion. (g) Find the work done by the string brake on the sliding ass. (h) Find the total work done on the syste consisting of the disk and the sliding ass. A B Figure P.46 what is its speed when it is farthest fro the Sun? The angular oentu of the coet about the Sun is conserved because no torque acts on the coet. The gravitational force eerted by the Sun on the coet has a oent ar of zero. 49. A constant horizontal force F is applied to a lawn roller having the for of a unifor solid cylinder of radius R and ass M (Fig. P.49). If the roller rolls without slipping on the horizontal surface, show that (a) the acceleration of the center of ass is 2F/3M and that (b) the iniu coefficient of friction necessary to prevent slipping is F/3Mg. (Hint: Consider the torque with respect to the center of ass.) R M F The onkey clibs the rope in an attept to reach the bananas. (a) Treating the syste as consisting of the onkey, bananas, rope, and pulley, evaluate the net torque about the pulley ais. (b) Using the results to part (a), deterine the total angular oentu about the pulley ais and describe the otion of the syste. Will the onkey reach the bananas? 5. A solid sphere of ass and radius r rolls without slipping along the track shown in Figure P.5. The sphere starts fro rest with its lowest point at height h above the botto of a loop of radius R, which is uch larger than r. (a) What is the iniu value that h can have (in ters of R) if the sphere is to coplete the loop? (b) What are the force coponents on the sphere at point P if h 3R? r r θ 46. A 00-kg unifor horizontal disk of radius 5.50 turns without friction at 2.50 rev/s on a vertical ais through its center, as shown in Figure P.46. A feedback echanis senses the angular speed of the disk, and a drive otor at A ensures that the angular speed reains constant. While the disk turns, a.20-kg ass on top of the disk slides outward in a radial slot. The.20-kg ass starts at the center of the disk at tie t 0 and oves outward with a constant speed of.25 c/s relative to the disk until it reaches the edge at t 440 s. The sliding ass eperiences no friction. Its otion is constrained by a brake at B so that its radial speed reains constant. The constraint produces tension in a light string tied to the ass. (a) Find the torque as a function of tie that the drive otor ust provide while the ass is sliding. (b) Find the value of this torque at t 440 s, just before the sliding ass finishes its otion. (c) Find the power that the drive otor ust deliver as a function of tie. (d) Find the value of the power when the sliding ass is just reaching the end of the slot. (e) Find the string tension as a function of R Figure P A string is wound around a unifor disk of radius R and ass M. The disk is released fro rest when the string is vertical and its top end is tied to a fied bar (Fig. P.47). Show that (a) the tension in the string is one-third the weight of the disk, (b) the agnitude of the acceleration of the center of ass is 2g/3, and (c) the speed of the center of ass is (4gh/3) /2 as the disk descends. Verify your answer to part (c) using the energy approach. h 48. Coet Halley oves about the Sun in an elliptical orbit, with its closest approach to the Sun being about AU and its greatest distance fro the Sun being 35.0 AU ( AU the average Earth Sun distance). If the coet s speed at its closest approach is 54.0 k/s, R Figure P.47 M Figure P A light rope passes over a light, frictionless pulley. A bunch of bananas of ass M is fastened at one end, and a onkey of ass M clings to the other (Fig. P.50). M Figure P.50 M h Figure P A thin rod with a ass of kg and a length of.24 is at rest, hanging vertically fro a strong fied hinge at its top end. Suddenly, a horizontal ipulsive force (4.7i) N is applied to it. (a) Suppose that the force acts at the botto end of the rod. Find the acceleration of the rod s center of ass and the horizontal force that the hinge eerts. (b) Suppose that the force acts at the idpoint of the rod. Find the acceleration of this point and the horizontal hinge reaction. (c) Where can the ipulse be applied so that the hinge eerts no horizontal force? (This point is called the center of percussion.) 53. At one oent, a bowling ball is both sliding and spinning on a horizontal surface such that its rotational kinetic energy equals its translational kinetic energy. Let v CM represent the ball s center-of-ass speed relative to the surface. Let v r represent the speed of the topost point on the ball s surface relative to the center of ass. Find the ratio v CM /v r. 54. A projectile of ass oves to the right with speed v i (Fig. P.54a). The projectile strikes and sticks to the end of a stationary rod of ass M and length d that is pivoted about a frictionless ale through its center (Fig. P.54b). (a) Find the angular speed of the syste right after the collision. (b) Deterine the fractional loss in echanical energy due to the collision. R P

57 Probles CHAPTER Rolling Motion and Angular Moentu (a) v i 55. A ass is attached to a cord passing through a sall hole in a frictionless, horizontal surface (Fig. P.55). The ass is initially orbiting with speed v i in a circle of radius r i. The cord is then slowly pulled fro below, and the radius of the circle decreases to r. (a) What is the speed of the ass when the radius is r? (b) Find the tension in the cord as a function of r. (c) How uch work W is done in oving fro r i to r? (Note: The tension depends on r.) (d) btain nuerical values for v, T, and W when r 0.00, 50.0 g, r i 0.300, and v i.50 /s. d Figure P.54 r i v i (b) ω cal grape at the top of his bald head, which itself has the shape of a sphere. After all of the children have had tie to giggle, the grape starts fro rest and rolls down your uncle s head without slipping. It loses contact with your uncle s scalp when the radial line joining it to the center of curvature akes an angle with the vertical. What is the easure of angle? 58. A thin rod of length h and ass M is held vertically with its lower end resting on a frictionless horizontal surface. The rod is then released to fall freely. (a) Deterine the speed of its center of ass just before it hits the horizontal surface. (b) Now suppose that the rod has a fied pivot at its lower end. Deterine the speed of the rod s center of ass just before it hits the surface. WEB 59. Two astronauts (Fig. P.59), each having a ass of 75.0 kg, are connected by a 0.0- rope of negligible ass. They are isolated in space, orbiting their center of ass at speeds of 5.00 /s. (a) Treating the astronauts as particles, calculate the agnitude of the angular oentu and (b) the rotational energy of the syste. By pulling on the rope, one of the astronauts shortens the distance between the to (c) What is the new angular oentu of the syste? (d) What are the astronauts new speeds? (e) What is the new rotational energy of the syste? (f) How uch work is done by the astronaut in shortening the rope? 60. Two astronauts (see Fig. P.59), each having a ass M, are connected by a rope of length d having negligible ass. They are isolated in space, orbiting their center of ass at speeds v. Treating the astronauts as particles, calculate (a) the agnitude of the angular oentu and (b) the rotational energy of the syste. By pulling on the rope, one of the astronauts shortens the distance between the to d/2. (c) What is the new angular oentu of the syste? (d) What are the astronauts new speeds? (e) What is the new rotational energy of the syste? (f) How uch work is done by the astronaut in shortening the rope? tate about an ais AB (Fig. P.6). A bullet of ass and speed v is shot at the face opposite ABCD at a height of 4a/3. The bullet becoes ebedded in the cube. Find the iniu value of v required to tip the cube so that it falls on face ABCD. Assue V M. v 2a 4a/3 Figure P A large, cylindrical roll of paper of initial radius R lies on a long, horizontal surface with the open end of the paper nailed to the surface. The roll is given a slight shove (v i 0) and begins to unroll. (a) Deterine the speed of the center of ass of the roll when its radius has diinished to r. (b) Calculate a nuerical value for this speed at r.00, assuing R (c) What happens to the energy of the syste when the paper is copletely unrolled? (Hint: Assue that the roll has a unifor density and apply energy ethods.) 63. A spool of wire of ass M and radius R is unwound under a constant force F (Fig. P.63). Assuing that the spool is a unifor solid cylinder that does not slip, show that (a) the acceleration of the center of ass is 4F/3M and that (b) the force of friction is to the right and is equal in agnitude to F/3. (c) If the cylinder starts fro rest and rolls without slipping, what is the speed of its center of ass after it has rolled through a distance d? A D C B Figure P.64 Probles 64 and 65. a horizontal surface and released, as shown in Figure P.64. (a) What is the angular speed of the disk once pure rolling takes place? (b) Find the fractional loss in kinetic energy fro the tie the disk is released until the tie pure rolling occurs. (Hint: Consider torques about the center of ass.) 65. Suppose a solid disk of radius R is given an angular speed i about an ais through its center and is then lowered to a horizontal surface and released, as shown in Proble 64 (see Fig. P.64). Furtherore, assue that the coefficient of friction between the disk and the surface is. (a) Show that the tie it takes for pure rolling otion to occur is R i /3g. (b) Show that the distance the disk travels before pure rolling occurs is R 2i 2 /8g. 66. A solid cube of side 2a and ass M is sliding on a frictionless surface with unifor velocity v, as shown in Figure P.66a. It hits a sall obstacle at the end of the table; this causes the cube to tilt, as shown in Figure F ω Figure P.55 CM 56. A bowler releases a bowling ball with no spin, sending it sliding straight down the alley toward the pins. The ball continues to slide for soe distance before its otion becoes rolling without slipping; of what order of agnitude is this distance? State the quantities you take as data, the values you easure or estiate for the, and your reasoning. 57. Following Thanksgiving dinner, your uncle falls into a deep sleep while sitting straight up and facing the television set. A naughty grandchild balances a sall spherid Figure P.59 Probles 59 and A solid cube of wood of side 2a and ass M is resting on a horizontal surface. The cube is constrained to ro- M R Figure P A unifor solid disk is set into rotation with an angular speed i about an ais through its center. While still rotating at this speed, the disk is placed into contact with 2a M v (a) Figure P.66 (b) Mg

58 Probles CHAPTER Rolling Motion and Angular Moentu P.66b. Find the iniu value of v such that the cube falls off the table. Note that the oent of inertia of the cube about an ais along one of its edges is 8Ma 2 /3. (Hint: The cube undergoes an inelastic collision at the edge.) 67. A plank with a ass M 6.00 kg rides on top of two identical solid cylindrical rollers that have R 5.00 c and 2.00 kg (Fig. P.67). The plank is pulled by a constant horizontal force of agnitude F 6.00 N applied to the end of the plank and perpendicular to the aes of the cylinders (which are parallel). The cylinders roll without slipping on a flat surface. Also, no slipping occurs between the cylinders and the plank. (a) Find the acceleration of the plank and that of the rollers. (b) What frictional forces are acting? M R R F that the critical angle for which the spool does not slip and reains stationary is cos c r R (Hint: At the critical angle, the line of action of the applied force passes through the contact point.) 70. In a deonstration that eploys a ballistics cart, a ball is projected vertically upward fro a cart oving with constant velocity along the horizontal direction. The ball lands in the catching cup of the cart because both the cart and the ball have the sae horizontal coponent of velocity. Now consider a ballistics cart on an incline aking an angle with the horizontal, as shown in Figure P.70. The cart (including its wheels) has a ass M, and the oent of inertia of each of the two wheels is R 2 /2. (a) Using conservation of energy considerations (assuing that there is no friction between the cart and the ales) and assuing pure rolling otion (that is, no slipping), show that the acceleration of the cart along the incline is ANSWERS T QUICK QUIZZES. There is very little resistance to otion that can reduce the kinetic energy of the rolling ball. Even though there is friction between the ball and the floor (if there were not, then no rotation would occur, and the ball would slide), there is no relative otion of the two surfaces (by the definition of rolling ), and so kinetic friction cannot reduce K. (Air drag and friction associated with deforation of the ball eventually stop the ball.).2 The bo. Because none of the bo s initial potential energy is converted to rotational kinetic energy, at any tie t 0 its translational kinetic energy is greater than that of the rolling ball..3 Zero. If she were oving directly toward the pole, r and p would be antiparallel to each other, and the sine of the angle between the is zero; therefore, L 0..4 Both (a) and (b) are false. The net force is not necessarily zero. If the line of action of the net force passes through the point, then the net torque about an ais passing through that point is zero even though the net force is not zero. Because the net force is not necessarily zero, you cannot conclude that the particle s velocity is constant..5 The student does work as he walks fro the ri of the platfor toward its center..6 Because it is perpendicular to the precessional otion of the top, the force of gravity does no work. This is a situation in which a force causes otion but does no work. Figure P A spool of wire rests on a horizontal surface as in Figure P.68. As the wire is pulled, the spool does not slip at the contact point P. n separate trials, each one of the forces F, F 2, F 3, and F 4 is applied to the spool. For each one of these forces, deterine the direction in which the spool will roll. Note that the line of action of F 2 passes through P. F 3 F 2 a M M 2 g sin (b) Note that the coponent of acceleration of the ball released by the cart is g sin. Thus, the coponent of the cart s acceleration is saller than that of the ball by the factor M/(M 2). Use this fact and kineatic equations to show that the ball overshoots the cart by an aount, where 4 M 2 sin cos v 2 yi 2 g and v yi is the initial speed of the ball iparted to it by the spring in the cart. (c) Show that the distance d that the ball travels easured along the incline is d 2v yi 2 sin g cos 2 F 4 R r θc F y P Figure P.68 Probles 68 and The spool of wire shown in Figure P.68 has an inner radius r and an outer radius R. The angle between the applied force and the horizontal can be varied. Show Figure P.70 θ

59 2.2 This is the Nearest ne Head 36 P U Z Z L E R This one-bottle wine holder is an interesting eaple of a echanical syste that sees to defy gravity. The syste (holder plus bottle) is balanced when its center of gravity is directly over the lowest support point. What two conditions are necessary for an object to ehibit this kind of stability? (Charles D. Winters) 362 CHAPTER 2 Static Equilibriu and Elasticity In Chapters 0 and we studied the dynaics of rigid objects that is, objects whose parts reain at a fied separation with respect to each other when subjected to eternal forces. Part of this chapter addresses the conditions under which a rigid object is in equilibriu. The ter equilibriu iplies either that the object is at rest or that its center of ass oves with constant velocity. We deal here only with the forer case, in which the object is described as being in static equilibriu. Static equilibriu represents a coon situation in engineering practice, and the principles it involves are of special interest to civil engineers, architects, and echanical engineers. If you are an engineering student you will undoubtedly take an advanced course in statics in the future. The last section of this chapter deals with how objects defor under load conditions. Such deforations are usually elastic and do not affect the conditions for equilibriu. An elastic object returns to its original shape when the deforing forces are reoved. Several elastic constants are defined, each corresponding to a different type of deforation. 2. THE CNDITINS FR EQUILIBRIUM Static Equilibriu and Elasticity Chapter utline 2. The Conditions for Equilibriu 2.2 More on the Center of Gravity 2.3 Eaples of Rigid bjects in Static Equilibriu 2.4 Elastic Properties of Solids c h a p t e r F θ P r d Figure 2. A single force F acts on a rigid object at the point P. Equivalent forces In Chapter 5 we stated that one necessary condition for equilibriu is that the net force acting on an object be zero. If the object is treated as a particle, then this is the only condition that ust be satisfied for equilibriu. The situation with real (etended) objects is ore cople, however, because these objects cannot be treated as particles. For an etended object to be in static equilibriu, a second condition ust be satisfied. This second condition involves the net torque acting on the etended object. Note that equilibriu does not require the absence of otion. For eaple, a rotating object can have constant angular velocity and still be in equilibriu. Consider a single force F acting on a rigid object, as shown in Figure 2.. The effect of the force depends on its point of application P. If r is the position vector of this point relative to, the torque associated with the force F about is given by Equation.7: r F Recall fro the discussion of the vector product in Section.2 that the vector is perpendicular to the plane fored by r and F. You can use the right-hand rule to deterine the direction of : Curl the fingers of your right hand in the direction of rotation that F tends to cause about an ais through : your thub then points in the direction of. Hence, in Figure 2. is directed toward you out of the page. As you can see fro Figure 2., the tendency of F to rotate the object about an ais through depends on the oent ar d, as well as on the agnitude of F. Recall that the agnitude of is Fd (see Eq. 0.9). Now suppose a rigid object is acted on first by force F and later by force F 2. If the two forces have the sae agnitude, they will produce the sae effect on the object only if they have the sae direction and the sae line of action. In other words, two forces F and F 2 are equivalent if and only if F F 2 and if and only if the two produce the sae torque about any ais. The two forces shown in Figure 2.2 are equal in agnitude and opposite in direction. They are not equivalent. The force directed to the right tends to rotate 36

60 2. The Conditions for Equilibriu CHAPTER 2 Static Equilibriu and Elasticity the object clockwise about an ais perpendicular to the diagra through, whereas the force directed to the left tends to rotate it counterclockwise about that ais. Suppose an object is pivoted about an ais through its center of ass, as shown in Figure 2.3. Two forces of equal agnitude act in opposite directions along parallel lines of action. A pair of forces acting in this anner for what is called a couple. (The two forces shown in Figure 2.2 also for a couple.) Do not ake the istake of thinking that the forces in a couple are a result of Newton s third law. They cannot be third-law forces because they act on the sae object. Third-law force pairs act on different objects. Because each force produces the sae torque Fd, the net torque has a agnitude of 2Fd. Clearly, the object rotates clockwise and undergoes an angular acceleration about the ais. With respect to rotational otion, this is a nonequilibriu situation. The net torque on the object gives rise to an angular acceleration according to the relationship 2Fd I (see Eq. 0.2). In general, an object is in rotational equilibriu only if its angular acceleration 0. Because I for rotation about a fied ais, our second necessary condition for equilibriu is that the net torque about any ais ust be zero. We now have two necessary conditions for equilibriu of an object: F 2 F Figure 2.2 The forces F and F 2 are not equivalent because they do not produce the sae torque about soe ais, even though they are equal in agnitude and opposite in direction. F 2 F F 3 r r r r F4 Figure 2.4 Construction showing that if the net torque is zero about origin, it is also zero about any other origin, such as. Regardless of the nuber of forces that are acting, if an object is in translational equilibriu and if the net torque is zero about one ais, then the net torque ust also be zero about any other ais. The point can be inside or outside the boundaries of the object. Consider an object being acted on by several forces such that the resultant force F F F 2 F 3 0. Figure 2.4 describes this situation (for clarity, only four forces are shown). The point of application of F relative to is specified by the position vector r. Siilarly, the points of application of F 2, F 3,...are specified by r 2, r 3,...(not shown). The net torque about an ais through is r F r 2 F 2 r 3 F 3 Now consider another arbitrary point having a position vector r relative to. The point of application of F relative to is identified by the vector r r. Likewise, the point of application of F 2 relative to is r 2 r, and so forth. Therefore, the torque about an ais through is (r r) F (r 2 r) F 2 (r 3 r) F 3 r F r 2 F 2 r 3 F 3 r (F F 2 F 3 ). The resultant eternal force ust equal zero. F 0 (2.) 2. The resultant eternal torque about any ais ust be zero. (2.2) The first condition is a stateent of translational equilibriu; it tells us that the linear acceleration of the center of ass of the object ust be zero when viewed fro an inertial reference frae. The second condition is a stateent of rotational equilibriu and tells us that the angular acceleration about any ais ust be zero. In the special case of static equilibriu, which is the ain subject of this chapter, the object is at rest and so has no linear or angular speed (that is, v CM 0 and 0). Quick Quiz 2. 0 (a) Is it possible for a situation to eist in which Equation 2. is satisfied while Equation 2.2 is not? (b) Can Equation 2.2 be satisfied while Equation 2. is not? The two vector epressions given by Equations 2. and 2.2 are equivalent, in general, to si scalar equations: three fro the first condition for equilibriu, and three fro the second (corresponding to, y, and z coponents). Hence, in a cople syste involving several forces acting in various directions, you could be faced with solving a set of equations with any unknowns. Here, we restrict our discussion to situations in which all the forces lie in the y plane. (Forces whose vector representations are in the sae plane are said to be coplanar.) With this restriction, we ust deal with only three scalar equations. Two of these coe fro balancing the forces in the and y directions. The third coes fro the torque equation naely, that the net torque about any point in the y plane ust be zero. Hence, the two conditions of equilibriu provide the equations F 0 F y 0 z 0 (2.3) where the ais of the torque equation is arbitrary, as we now show. Conditions for equilibriu F d d CM Figure 2.3 Two forces of equal agnitude for a couple if their lines of action are different parallel lines. In this case, the object rotates clockwise. The net torque about any ais is 2Fd. F y,y 2,y 2 2 CM 3,y 3 3 Figure 2.5 An object can be divided into any sall particles each having a specific ass and specific coordinates. These particles can be used to locate the center of ass. Because the net force is assued to be zero (given that the object is in translational equilibriu), the last ter vanishes, and we see that the torque about is equal to the torque about. Hence, if an object is in translational equilibriu and the net torque is zero about one point, then the net torque ust be zero about any other point. 2.2 MRE N THE CENTER F GRAVITY We have seen that the point at which a force is applied can be critical in deterining how an object responds to that force. For eaple, two equal-agnitude but oppositely directed forces result in equilibriu if they are applied at the sae point on an object. However, if the point of application of one of the forces is oved, so that the two forces no longer act along the sae line of action, then a force couple results and the object undergoes an angular acceleration. (This is the situation shown in Figure 2.3.) Whenever we deal with a rigid object, one of the forces we ust consider is the force of gravity acting on it, and we ust know the point of application of this force. As we learned in Section 9.6, on every object is a special point called its center of gravity. All the various gravitational forces acting on all the various ass eleents of the object are equivalent to a single gravitational force acting through this point. Thus, to copute the torque due to the gravitational force on an object of ass M, we need only consider the force Mg acting at the center of gravity of the object. How do we find this special point? As we entioned in Section 9.6, if we assue that g is unifor over the object, then the center of gravity of the object coincides with its center of ass. To see that this is so, consider an object of arbitrary shape lying in the y plane, as illustrated in Figure 2.5. Suppose the object is divided into a large nuber of particles of asses, 2, 3,... having coordinates (, y ), ( 2, y 2 ), ( 3, y 3 ),.... In

61 2.3 Eaples of Rigid bjects in Static Equilibriu CHAPTER 2 Static Equilibriu and Elasticity Equation 9.28 we defined the coordinate of the center of ass of such an object to be CM We use a siilar equation to define the y coordinate of the center of ass, replacing each with its y counterpart. Let us now eaine the situation fro another point of view by considering the force of gravity eerted on each particle, as shown in Figure 2.6. Each particle contributes a torque about the origin equal in agnitude to the particle s weight g ultiplied by its oent ar. For eaple, the torque due to the force g is g, where g is the agnitude of the gravitational field at the position of the particle of ass. We wish to locate the center of gravity, the point at which application of the single gravitational force Mg (where M is the total ass of the object) has the sae effect on rotation as does the cobined effect of all the individual gravitational forces i g i. Equating the torque resulting fro Mg acting at the center of gravity to the su of the torques acting on the individual particles gives ( g 2 g 2 3 g 3 ) CG g 2 g g 3 3 This epression accounts for the fact that the gravitational field strength g can in general vary over the object. If we assue unifor g over the object (as is usually the case), then the g ters cancel and we obtain CG i i i i i (2.4) y,y g CG 2,y 2 2 g 3,y 3 F g = Mg 3 g Figure 2.6 The center of gravity of an object is located at the center of ass if g is constant over the object. Proble-Solving Hints bjects in Static Equilibriu Draw a siple, neat diagra of the syste. Isolate the object being analyzed. Draw a free-body diagra and then show and label all eternal forces acting on the object, indicating where those forces are applied. Do not include forces eerted by the object on its surroundings. (For systes that contain ore than one object, draw a separate free-body diagra for each one.) Try to guess the correct direction for each force. If the direction you select leads to a negative force, do not be alared; this erely eans that the direction of the force is the opposite of what you guessed. Establish a convenient coordinate syste for the object and find the coponents of the forces along the two aes. Then apply the first condition for equilibriu. Reeber to keep track of the signs of all force coponents. Choose a convenient ais for calculating the net torque on the object. Reeber that the choice of origin for the torque equation is arbitrary; therefore, choose an origin that siplifies your calculation as uch as possible. Note that a force that acts along a line passing through the point chosen as the origin gives zero contribution to the torque and thus can be ignored. The first and second conditions for equilibriu give a set of linear equations containing several unknowns, and these equations can be solved siultaneously. Coparing this result with Equation 9.28, we see that the center of gravity is located at the center of ass as long as the object is in a unifor gravitational field. In several eaples presented in the net section, we are concerned with hoogeneous, syetric objects. The center of gravity for any such object coincides with its geoetric center. 2.3 EXAMPLES F RIGID BJECTS IN STATIC EQUILIBRIUM The photograph of the one-bottle wine holder on the first page of this chapter shows one eaple of a balanced echanical syste that sees to defy gravity. For the syste (wine holder plus bottle) to be in equilibriu, the net eternal force ust be zero (see Eq. 2.) and the net eternal torque ust be zero (see Eq. 2.2). The second condition can be satisfied only when the center of gravity of the syste is directly over the support point. In working static equilibriu probles, it is iportant to recognize all the eternal forces acting on the object. Failure to do so results in an incorrect analysis. When analyzing an object in equilibriu under the action of several eternal forces, use the following procedure. A large balanced rock at the Garden of the Gods in Colorado Springs, Colorado an eaple of stable equilibriu. EXAMPLE 2. The Seesaw A unifor 40.0-N board supports a father and daughter weighing 800 N and 350 N, respectively, as shown in Figure 2.7. If the support (called the fulcru) is under the center of gravity of the board and if the father is.00 fro the center, (a) deterine the agnitude of the upward force n eerted on the board by the support. First note that, in addition to n, the eternal forces acting on the board are the downward forces eerted by each person and the force of gravity acting on the board. We know that the board s center of gravity is at its geoetric center because we were told the board is unifor. Because the syste is in static equilibriu, the upward force n ust balance all the downward forces. Fro F y 0, we have, once we define upward as the positive y direction, n 800 N 350 N 40.0 N 0 n 90 N (The equation F 0 also applies, but we do not need to consider it because no forces act horizontally on the board.) (b) Deterine where the child should sit to balance the syste. To find this position, we ust invoke the second condition for equilibriu. Taking an ais perpendicular to the page through the center of gravity of the board as the ais for our torque equation (this eans that the torques 800 N.00 Figure 2.7 n 40.0 N A balanced syste. 350 N

62 2.3 Eaples of Rigid bjects in Static Equilibriu CHAPTER 2 Static Equilibriu and Elasticity produced by n and the force of gravity acting on the board about this ais are zero), we see fro that (c) Repeat part (b) for another ais. To illustrate that the choice of ais is arbitrary, let us choose an ais perpendicular to the page and passing Quick Quiz 2.2 In Eaple 2., if the fulcru did not lie under the board s center of gravity, what other inforation would you need to solve the proble? EXAMPLE 2.2 (800 N)(.00 ) (350 N) A Weighted Hand A person holds a 50.0-N sphere in his hand. The forear is horizontal, as shown in Figure 2.8a. The biceps uscle is attached 3.00 c fro the joint, and the sphere is 35.0 c fro the joint. Find the upward force eerted by the biceps on the forear and the downward force eerted by the upper ar on the forear and acting at the joint. Neglect the weight of the forear. Biceps Figure 2.8 d g = 50.0 N d = 3.00 c = 35.0 c g (a) The biceps uscle pulls upward with a force F that is essentially at right angles to the forear. (b) The echanical odel for the syste described in part (a). EXAMPLE 2.3 R F d g Standing on a Horizontal Bea A unifor horizontal bea with a length of 8.00 and a weight of 200 N is attached to a wall by a pin connection. Its far end is supported by a cable that akes an angle of 53.0 with through the location of the father. Recall that the sign of the torque associated with a force is positive if that force tends to rotate the syste counterclockwise, while the sign of the torque is negative if the force tends to rotate the syste clockwise. In this case, yields n(.00 ) (40.0 N)(.00 ) (350 N)(.00 ) 0 Fro part (a) we know that n 90 N. Thus, we can solve for to find This result is in agreeent with the one we obtained in part (b). 0 We siplify the situation by odeling the forear as a bar as shown in Figure 2.8b, where F is the upward force eerted by the biceps and R is the downward force eerted by the upper ar at the joint. Fro the first condition for equilibriu, we have, with upward as the positive y direction, () F y F R 50.0 N 0 Fro the second condition for equilibriu, we know that the su of the torques about any point ust be zero. With the joint as the ais, we have Fd g 0 F(3.00 c) (50.0 N)(35.0 c) 0 F This value for F can be substituted into Equation () to give R 533 N. As this eaple shows, the forces at joints and in uscles can be etreely large. Eercise In reality, the biceps akes an angle of 5.0 with the vertical; thus, F has both a vertical and a horizontal coponent. Find the agnitude of F and the coponents of R when you include this fact in your analysis. Answer F 604 N, R 56 N, R y 533 N. 583 N the horizontal (Fig. 2.9a). If a 600-N person stands 2.00 fro the wall, find the tension in the cable, as well as the agnitude and direction of the force eerted by the wall on the bea. Figure 2.9 R θ R sin θ R cos θ N 600 N 8.00 (a) (b) 200 N 200 N 53.0 First we ust identify all the eternal forces acting on the bea: They are the 200-N force of gravity, the force T eerted by the cable, the force R eerted by the wall at the pivot, and the 600-N force that the person eerts on the bea. These forces are all indicated in the free-body diagra for the bea shown in Figure 2.9b. When we consider directions for forces, it soeties is helpful if we iagine what would happen if a force were suddenly reoved. For eaple, if the wall were to vanish suddenly, T 53.0 T sin 53.0 T cos 53.0 (a) A unifor bea supported by a cable. (b) The free-body diagra for the bea. (c) The free-body diagra for the bea showing the coponents of R and T. the left end of the bea would probably ove to the left as it begins to fall. This tells us that the wall is not only holding the bea up but is also pressing outward against it. Thus, we draw the vector R as shown in Figure 2.9b. If we resolve T and R into horizontal and vertical coponents, as shown in Figure 2.9c, and apply the first condition for equilibriu, we obtain () F R cos T cos (2) where we have chosen rightward and upward as our positive directions. Because R, T, and are all unknown, we cannot obtain a solution fro these epressions alone. (The nuber of siultaneous equations ust equal the nuber of unknowns for us to be able to solve for the unknowns.) Now let us invoke the condition for rotational equilibriu. A convenient ais to choose for our torque equation is the one that passes through the pin connection. The feature that akes this point so convenient is that the force R and the horizontal coponent of T both have a oent ar of zero; hence, these forces provide no torque about this point. Recalling our counterclockwise-equals-positive convention for the sign of the torque about an ais and noting that the oent ars of the 600-N, 200-N, and T sin 53.0 forces are 2.00, 4.00, and 8.00, respectively, we obtain (T sin 53.0)(8.00 ) T F y R sin T sin N 200 N 0 (600 N)(2.00 ) (200 N )(4.00 ) 0 33 N Thus, the torque equation with this ais gives us one of the unknowns directly! We now substitute this value into Equations () and (2) and find that R cos 88 N R sin 550 N We divide the second equation by the first and, recalling the trigonoetric identity sin /cos tan, we obtain tan 550 N N 7. This positive value indicates that our estiate of the direction of R was accurate. Finally, R 88 N 88 N cos cos N If we had selected soe other ais for the torque equation, the solution would have been the sae. For eaple, if

63 2.3 Eaples of Rigid bjects in Static Equilibriu CHAPTER 2 Static Equilibriu and Elasticity we had chosen an ais through the center of gravity of the bea, the torque equation would involve both T and R. However, this equation, coupled with Equations () and (2), could still be solved for the unknowns. Try it! When any forces are involved in a proble of this nature, it is convenient to set up a table. For instance, for the eaple just given, we could construct the following table. Setting the su of the ters in the last colun equal to zero represents the condition of rotational equilibriu. EXAMPLE 2.4 The Leaning Ladder A unifor ladder of length and weight g 50 N rests against a sooth, vertical wall (Fig. 2.0a). If the coefficient of static friction between the ladder and the ground is s 0.40, find the iniu angle in at which the ladder does not slip. The free-body diagra showing all the eternal forces acting on the ladder is illustrated in Figure 2.0b. The reaction force R eerted by the ground on the ladder is the vector su of a noral force n and the force of static friction f s. The reaction force P eerted by the wall on the ladder is horizontal because the wall is frictionless. Notice how we have included only forces that act on the ladder. For eaple, the forces eerted by the ladder on the ground and on the wall are not part of the proble and thus do not appear in the free-body diagra. Applying the first condition θ Figure 2.0 (a) (a) A unifor ladder at rest, leaning against a sooth wall. The ground is rough. (b) The free-body diagra for the ladder. Note that the forces R, g, and P pass through a coon point. n φ f R θ (b) g P Moent Ar Force Relative to Torque About Coponent () (N) T sin (8.00)T sin 53.0 T cos N 4.00 (4.00)(200) 600 N 2.00 (2.00)(600) R sin 0 0 R cos 0 0 for equilibriu to the ladder, we have F f P 0 F y n g 0 Fro the second equation we see that n g 50 N. Furtherore, when the ladder is on the verge of slipping, the force of friction ust be a aiu, which is given by f s,a sn 0.40(50 N) 20 N. (Recall Eq. 5.8: f s s n.) Thus, at this angle, P 20 N. To find in, we ust use the second condition for equilibriu. When we take the torques about an ais through the origin at the botto of the ladder, we have P sin g 2 cos 0 Because P 20 N when the ladder is about to slip, and because g 50 N, this epression gives tan in g 50 N.25 2P 40 N in 5 An alternative approach is to consider the intersection of the lines of action of forces g and P. Because the torque about any origin ust be zero, the torque about ust be zero. This requires that the line of action of R (the resultant of n and f ) pass through. In other words, because the ladder is stationary, the three forces acting on it ust all pass through soe coon point. (We say that such forces are concurrent.) With this condition, you could then obtain the angle that R akes with the horizontal (where is greater than ). Because this approach depends on the length of the ladder, you would have to know the value of to obtain a value for in. Eercise For the angles labeled in Figure 2.0, show that tan 2 tan. EXAMPLE 2.5 Negotiating a Curb (a) Estiate the agnitude of the force F a person ust apply to a wheelchair s ain wheel to roll up over a sidewalk curb (Fig. 2.a). This ain wheel, which is the one that coes in contact with the curb, has a radius r, and the height of the curb is h. Norally, the person s hands supply the required force to a slightly saller wheel that is concentric with the ain wheel. We assue that the radius of the saller wheel is the sae as the radius of the ain wheel, and so we can use r for our radius. Let us estiate a cobined weight of g 400 N for the person and the wheelchair and choose a wheel radius of r 30 c, as shown in Figure 2.b. We also pick a curb height of h 0 c. We assue that the wheelchair and occupant are syetric, and that each wheel supports a weight of 700 N. We then proceed to analyze only one of the wheels. When the wheel is just about to be raised fro the street, the reaction force eerted by the ground on the wheel at point Q goes to zero. Hence, at this tie only three forces act on the wheel, as shown in Figure 2.c. However, the force R, which is the force eerted on the wheel by the curb, acts at point P, and so if we choose to have our ais of rotation pass through point P, we do not need to include R in our torque equation. Fro the triangle PQ shown in Figure 2.b, we see that the oent ar d of the gravitational force g acting on the wheel relative to point P is d!r 2 (r h) 2!2rh h 2 The oent ar of F relative to point P is 2r h. Therefore, the net torque acting on the wheel about point P is gd F(2r h) 0 g!2rh h 2 F(2r h) 0 g!2rh h2 F 2r h (700 N)!2(0.3 )(0. ) (0. )2 F 300 N 2(0.3 ) 0. (Notice that we have kept only one digit as significant.) This result indicates that the force that ust be applied to each wheel is substantial. You ay want to estiate the force required to roll a wheelchair up a typical sidewalk accessibility rap for coparison. (b) Deterine the agnitude and direction of R. We use the first condition for equilibriu to deterine the direction: F F R cos 0 F y R sin g 0 Dividing the second equation by the first gives tan g 700 N ; 70 F 300 N Figure 2. r h g Q C θ (a) r θ d (b) g (c) F (d) R R F F P P h 2r h (a) A wheelchair and person of total weight g being raised over a curb by a force F. (b) Details of the wheel and curb. (c) The free-body diagra for the wheel when it is just about to be raised. Three forces act on the wheel at this instant: F, which is eerted by the hand; R, which is eerted by the curb; and the gravitational force g. (d) The vector su of the three eternal forces acting on the wheel is zero.

64 2.3 Eaples of Rigid bjects in Static Equilibriu CHAPTER 2 Static Equilibriu and Elasticity APPLICATIN Analysis of a Truss Roofs, bridges, and other structures that ust be both strong and lightweight often are ade of trusses siilar to the one shown in Figure 2.2a. Iagine that this truss structure represents part of a bridge. To approach this proble, we assue that the structural coponents are connected by pin joints. We also assue that the entire structure is free to slide horizontally because it sits on rockers on each end, which allow it to ove back and forth as it undergoes theral epansion and contraction. Assuing the ass of the bridge structure is negligible copared with the load, let us calculate the forces of tension or copression in all the structural coponents when it is supporting a N load at the center (see Proble 58). The force notation that we use here is not of our usual forat. Until now, we have used the notation F AB to ean the force eerted by A on B. For this application, however, all double-letter subscripts on F indicate only the body eerting the force. The body on which a given force acts is not naed in the subscript. For eaple, in Figure 2.2, F AB is the force eerted by strut AB on the pin at A. First, we apply Newton s second law to the truss as a whole in the vertical direction. Internal forces do not enter into this accounting. We balance the weight of the load with the noral forces eerted at the two ends by the supports on which the bridge rests: A We can use the right triangle shown in Figure 2.d to obtain R : A FAB n A R!(g) 2 F 2!(700 N) 2 (300 N) F AC Figure 2.2 B B F BC 30 F CA Load: N C 50 (a) C (b) D 800 N (a) Truss structure for a bridge. (b) The forces acting on the pins at points A, C, and E. As an eercise, you should diagra the forces acting on the pin at point B. F g F DC D F CE F EC n E E E F ED Eercise Solve this proble by noting that the three forces acting on the wheel are concurrent (that is, that all three pass through the point C). The three forces for the sides of the triangle shown in Figure 2.d. F y n A n E F g 0 n A n E N Net, we calculate the torque about A, noting that the overall length of the bridge structure is L 50 : Ln E (L/2)F g 0 n E F g / N Although we could repeat the torque calculation for the right end (point E), it should be clear fro syetry arguents that n A N. Now let us balance the vertical forces acting on the pin at point A. If we assue that strut AB is in copression, then the force F AB that the strut eerts on the pin at point A has a negative y coponent. (If the strut is actually in tension, our calculations will result in a negative value for the agnitude of the force, still of the correct size): F y n A F AB sin 30 0 F AB N The positive result shows that our assuption of copression was correct. We can now find the forces acting in the strut between A and C by considering the horizontal forces acting on the pin at point A. Because point A is not accelerating, we can safely assue that F AC ust point toward the right (Fig. 2.2b); this indicates that the bar between points A and C is under tension: F F AC F AB cos 30 0 F AC (7 200 N)cos N Now let us consider the vertical forces acting on the pin at point C. We shall assue that strut BC is in tension. (Iagine the subsequent otion of the pin at point C if strut BC were to break suddenly.) n the basis of syetry, we assert that F BC F DC and that F AC F EC : F y 2 F BC sin N 0 F BC N Finally, we balance the horizontal forces on B, assuing that strut BD is in copression: F F AB cos 30 F BC cos 30 F BD 0 (7 200 N)cos 30 (7 200 N)cos 30 F BD 0 F BD N Thus, the top bar in a bridge of this design ust be very strong. 2.4 ELASTIC PRPERTIES F SLIDS In our study of echanics thus far, we have assued that objects reain undefored when eternal forces act on the. In reality, all objects are deforable. That is, it is possible to change the shape or the size of an object (or both) by applying eternal forces. As these changes take place, however, internal forces in the object resist the deforation. We shall discuss the deforation of solids in ters of the concepts of stress and strain. Stress is a quantity that is proportional to the force causing a deforation; ore specifically, stress is the eternal force acting on an object per unit cross-sectional area. Strain is a easure of the degree of deforation. It is found that, for sufficiently sall stresses, strain is proportional to stress; the constant of proportionality depends on the aterial being defored and on the nature of the deforation. We call this proportionality constant the elastic odulus. The elastic odulus is therefore the ratio of the stress to the resulting strain: Elastic odulus stress (2.5) strain In a very real sense it is a coparison of what is done to a solid object (a force is applied) and how that object responds (it defors to soe etent). A plastic odel of an arch structure under load conditions. The wavy lines indicate regions where the stresses are greatest. Such odels are useful in designing architectural coponents.

65 2.4 Elastic Properties of Solids CHAPTER 2 Static Equilibriu and Elasticity We consider three types of deforation and define an elastic odulus for each:. Young s odulus, which easures the resistance of a solid to a change in its length 2. Shear odulus, which easures the resistance to otion of the planes of a solid sliding past each other 3. Bulk odulus, which easures the resistance of solids or liquids to changes in their volue F h A (a) F Fied face tic liit, the object is peranently distorted and does not return to its original shape after the stress is reoved. Hence, the shape of the object is peranently changed. As the stress is increased even further, the aterial ultiately breaks. Quick Quiz 2.3 What is Young s odulus for the elastic solid whose stress strain curve is depicted in Figure 2.4? L i L F Quick Quiz 2.4 Young s Modulus: Elasticity in Length Consider a long bar of cross-sectional area A and initial length L i that is claped at one end, as in Figure 2.3. When an eternal force is applied perpendicular to the cross section, internal forces in the bar resist distortion ( stretching ), but the bar attains an equilibriu in which its length L f is greater than L i and in which the eternal force is eactly balanced by internal forces. In such a situation, the bar is said to be stressed. We define the tensile stress as the ratio of the agnitude of the eternal force F to the cross-sectional area A. The tensile strain in this case is defined as the ratio of the change in length L to the original length L i. We define Young s odulus by a cobination of these two ratios: tensile stress Y (2.6) tensile strain F/A L/L i Young s odulus is typically used to characterize a rod or wire stressed under either tension or copression. Note that because strain is a diensionless quantity, Y has units of force per unit area. Typical values are given in Table 2.. Eperients show (a) that for a fied applied force, the change in length is proportional to the original length and (b) that the force necessary to produce a given strain is proportional to the cross-sectional area. Both of these observations are in accord with Equation 2.6. The elastic liit of a substance is defined as the aiu stress that can be applied to the substance before it becoes peranently defored. It is possible to eceed the elastic liit of a substance by applying a sufficiently large stress, as seen in Figure 2.4. Initially, a stress strain curve is a straight line. As the stress increases, however, the curve is no longer straight. When the stress eceeds the elas- TABLE 2. Typical Values for Elastic Modulus Young s Modulus Shear Modulus Bulk Modulus Substance (N/ 2 ) (N/ 2 ) (N/ 2 ) Tungsten Steel Copper Brass Aluinu Glass Quartz Water Mercury Figure 2.3 A long bar claped at one end is stretched by an aount L under the action of a force F. Young s odulus Stress (MN/ 2 ) A Elastic Breaking liit point Elastic behavior Strain Figure 2.4 Stress-versus-strain curve for an elastic solid. F f s Shear odulus QuickLab Estiate the shear odulus for the pages of your tetbook. Does the thickness of the book have any effect on the odulus value? Bulk odulus (b) Figure 2.5 (a) A shear deforation in which a rectangular block is distorted by two forces of equal agnitude but opposite directions applied to two parallel faces. (b) A book under shear stress. A aterial is said to be ductile if it can be stressed well beyond its elastic liit without breaking. A brittle aterial is one that breaks soon after the elastic liit is reached. How would you classify the aterial in Figure 2.4? Shear Modulus: Elasticity of Shape Another type of deforation occurs when an object is subjected to a force tangential to one of its faces while the opposite face is held fied by another force (Fig. 2.5a). The stress in this case is called a shear stress. If the object is originally a rectangular block, a shear stress results in a shape whose cross-section is a parallelogra. A book pushed sideways, as shown in Figure 2.5b, is an eaple of an object subjected to a shear stress. To a first approiation (for sall distortions), no change in volue occurs with this deforation. We define the shear stress as F/A, the ratio of the tangential force to the area A of the face being sheared. The shear strain is defined as the ratio /h, where is the horizontal distance that the sheared face oves and h is the height of the object. In ters of these quantities, the shear odulus is shear stress S (2.7) shear strain F/A /h Values of the shear odulus for soe representative aterials are given in Table 2.. The unit of shear odulus is force per unit area. Bulk Modulus: Volue Elasticity Bulk odulus characterizes the response of a substance to unifor squeezing or to a reduction in pressure when the object is placed in a partial vacuu. Suppose that the eternal forces acting on an object are at right angles to all its faces, as shown in Figure 2.6, and that they are distributed uniforly over all the faces. As we shall see in Chapter 5, such a unifor distribution of forces occurs when an object is iersed in a fluid. An object subject to this type of deforation undergoes a change in volue but no change in shape. The volue stress is defined as the ratio of the agnitude of the noral force F to the area A. The quantity P F/A is called the pressure. If the pressure on an object changes by an aount P F/A, then the object will eperience a volue change V. The volue strain is equal to the change in volue V divided by the initial volue V i. Thus, fro Equation 2.5, we can characterize a volue ( bulk ) copression in ters of the bulk odulus, which is defined as B volue stress F/A P volue strain V/V i V/V i (2.8)

66 2.4 Elastic Properties of Solids CHAPTER 2 Static Equilibriu and Elasticity V i forces are released after the concrete cures; this results in a peranent tension in the steel and hence a copressive stress on the concrete. This enables the concrete slab to support a uch heavier load. F V i V Figure 2.6 When a solid is under unifor pressure, it undergoes a change in volue but no change in shape. This cube is copressed on all sides by forces noral to its si faces. A negative sign is inserted in this defining equation so that B is a positive nuber. This aneuver is necessary because an increase in pressure (positive P ) causes a decrease in volue (negative V ) and vice versa. Table 2. lists bulk oduli for soe aterials. If you look up such values in a different source, you often find that the reciprocal of the bulk odulus is listed. The reciprocal of the bulk odulus is called the copressibility of the aterial. Note fro Table 2. that both solids and liquids have a bulk odulus. However, no shear odulus and no Young s odulus are given for liquids because a liquid does not sustain a shearing stress or a tensile stress (it flows instead). Prestressed Concrete If the stress on a solid object eceeds a certain value, the object fractures. The aiu stress that can be applied before fracture occurs depends on the nature of the aterial and on the type of applied stress. For eaple, concrete has a tensile strength of about N/ 2, a copressive strength of N/ 2, and a shear strength of N/ 2. If the applied stress eceeds these values, the concrete fractures. It is coon practice to use large safety factors to prevent failure in concrete structures. Concrete is norally very brittle when it is cast in thin sections. Thus, concrete slabs tend to sag and crack at unsupported areas, as shown in Figure 2.7a. The slab can be strengthened by the use of steel rods to reinforce the concrete, as illustrated in Figure 2.7b. Because concrete is uch stronger under copression (squeezing) than under tension (stretching) or shear, vertical coluns of concrete can support very heavy loads, whereas horizontal beas of concrete tend to sag and crack. However, a significant increase in shear strength is achieved if the reinforced concrete is prestressed, as shown in Figure 2.7c. As the concrete is being poured, the steel rods are held under tension by eternal forces. The eternal Concrete Load force (a) Cracks (b) Steel reinforcing rod (c) Steel rod under tension Figure 2.7 (a) A concrete slab with no reinforceent tends to crack under a heavy load. (b) The strength of the concrete is increased by using steel reinforceent rods. (c) The concrete is further strengthened by prestressing it with steel rods under tension. QuickLab Support a new flat eraser (art gu or Pink Pearl will do) on two parallel pencils at least 3 c apart. Press down on the iddle of the top surface just enough to ake the top face of the eraser curve a bit. Is the top face under tension or copression? How about the botto? Why does a flat slab of concrete supported at the ends tend to crack on the botto face and not the top? EXAMPLE 2.6 EXAMPLE 2.7 Squeezing a Brass Sphere A solid brass sphere is initially surrounded by air, and the air pressure eerted on it is N/ 2 (noral atospheric pressure). The sphere is lowered into the ocean to a depth at which the pressure is N/ 2. The volue of the sphere in air is By how uch does this volue change once the sphere is suberged? Stage Design Recall Eaple 8.0, in which we analyzed a cable used to support an actor as he swung onto the stage. The tension in the cable was 940 N. What diaeter should a 0--long steel wire have if we do not want it to stretch ore than 0.5 c under these conditions? Fro the definition of Young s odulus, we can solve for the required cross-sectional area. Assuing that the cross section is circular, we can deterine the diaeter of the wire. Fro Equation 2.6, we have Y F/A L/L i A FL i Y L (940 N)(0 ) ( N/ 2 )(0.005 ) Fro the definition of bulk odulus, we have B P V/V i SUMMARY The radius of the wire can be found fro A r 2 : A r!! d 2r 2(.7 ) 3.4 To provide a large argin of safety, we would probably use a fleible cable ade up of any saller wires having a total cross-sectional area substantially greater than our calculated value. V V ip B Because the final pressure is so uch greater than the initial pressure, we can neglect the initial pressure and state that P P f P i P f N/ 2. Therefore, V ( )( N/ 2 ) N/ 2 The negative sign indicates a decrease in volue A rigid object is in equilibriu if and only if the resultant eternal force acting on it is zero and the resultant eternal torque on it is zero about any ais: F 0 (2.) (2.2) The first condition is the condition for translational equilibriu, and the second is the condition for rotational equilibriu. These two equations allow you to analyze a great variety of probles. Make sure you can identify forces unabiguously, create a free-body diagra, and then apply Equations 2. and 2.2 and solve for the unknowns. 0

67 Probles CHAPTER 2 Static Equilibriu and Elasticity The force of gravity eerted on an object can be considered as acting at a single point called the center of gravity. The center of gravity of an object coincides with its center of ass if the object is in a unifor gravitational field. We can describe the elastic properties of a substance using the concepts of stress and strain. Stress is a quantity proportional to the force producing a deforation; strain is a easure of the degree of deforation. Strain is proportional to stress, and the constant of proportionality is the elastic odulus: Elastic odulus stress (2.5) strain Three coon types of deforation are () the resistance of a solid to elongation under a load, characterized by Young s odulus Y; (2) the resistance of a solid to the otion of internal planes sliding past each other, characterized by the shear odulus S; and (3) the resistance of a solid or fluid to a volue change, characterized by the bulk odulus B. QUESTINS. Can a body be in equilibriu if only one eternal force acts on it? Eplain. 2. Can a body be in equilibriu if it is in otion? Eplain. 3. Locate the center of gravity for the following unifor objects: (a) sphere, (b) cube, (c) right circular cylinder. 4. The center of gravity of an object ay be located outside the object. Give a few eaples for which this is the case. 5. You are given an arbitrarily shaped piece of plywood, together with a haer, nail, and plub bob. How could you use these ites to locate the center of gravity of the plywood? (Hint: Use the nail to suspend the plywood.) 6. For a chair to be balanced on one leg, where ust the center of gravity of the chair be located? 7. Can an object be in equilibriu if the only torques acting on it produce clockwise rotation? 8. A tall crate and a short crate of equal ass are placed side by side on an incline (without touching each other). As the incline angle is increased, which crate will topple first? Eplain. 9. When lifting a heavy object, why is it recoended to PRBLEMS keep the back as vertical as possible, lifting fro the knees, rather than bending over and lifting fro the waist? 0. Give a few eaples in which several forces are acting on a syste in such a way that their su is zero but the syste is not in equilibriu.. If you easure the net torque and the net force on a syste to be zero, (a) could the syste still be rotating with respect to you? (b) Could it be translating with respect to you? 2. A ladder is resting inclined against a wall. Would you feel safer clibing up the ladder if you were told that the ground is frictionless but the wall is rough or that the wall is frictionless but the ground is rough? Justify your answer. 3. What kind of deforation does a cube of Jell- ehibit when it jiggles? 4. Ruins of ancient Greek teples often have intact vertical coluns, but few horizontal slabs of stone are still in place. Can you think of a reason why this is so?, 2, 3 = straightforward, interediate, challenging = full solution available in the Student s Manual and Study Guide WEB = solution posted at = Coputer useful in solving proble = Interactive Physics = paired nuerical/sybolic probles Section 2. The Conditions for Equilibriu. A baseball player holds a 36-oz bat (weight 0.0 N) with one hand at the point (Fig. P2.). The bat is in equilibriu. The weight of the bat acts along a line 60.0 c to the right of. Deterine the force and the torque eerted on the bat by the player c Figure P2. g WEB 2. Write the necessary conditions of equilibriu for the body shown in Figure P2.2. Take the origin of the torque equation at the point. R R y θ F g Figure P A unifor bea of ass b and length supports blocks of asses and 2 at two positions, as shown in Figure P2.3. The bea rests on two points. For what value of will the bea be balanced at P such that the noral force at is zero? 2 CG d P Figure P A student gets his car stuck in a snow drift. Not at a loss, having studied physics, he attaches one end of a stout rope to the vehicle and the other end to the trunk of a nearby tree, allowing for a very sall aount of slack. The student then eerts a force F on the center of the rope in the direction perpendicular to the car tree line, as shown in Figure P2.4. If the rope is inetensible and if the agnitude of the applied force is 500 N, what is the force on the car? (Assue equilibriu conditions.) 2 Figure P2.4 F F y 2 F 0.50 Tree Section 2.2 WEB More on the Center of Gravity 5. A 3.00-kg particle is located on the ais at 5.00, and a 4.00-kg particle is located on the ais at Find the center of gravity of this twoparticle syste. 6. A circular pizza of radius R has a circular piece of radius R/2 reoved fro one side, as shown in Figure P2.6. Clearly, the center of gravity has oved fro C to C along the ais. Show that the distance fro C to C is R/6. (Assue that the thickness and density of the pizza are unifor throughout.) 8.0 c 4.0 c C 7. A carpenter s square has the shape of an L, as shown in Figure P2.7. Locate its center of gravity. 2.0 c C Figure P2.6 Figure P c 8. Pat builds a track for his odel car out of wood, as illustrated in Figure P2.8. The track is 5.00 c wide,.00 high, and 3.00 long, and it is solid. The runway is cut so that it fors a parabola described by the equation y ( 3) 2 /9. Locate the horizontal position of the center of gravity of this track. 9. Consider the following ass distribution: 5.00 kg at (0, 0), 3.00 kg at (0, 4.00), and 4.00 kg at (3.00, 0). Where should a fourth ass of 8.00 kg be placed so that the center of gravity of the four-ass arrangeent will be at (0, 0)?

68 Probles CHAPTER 2 Static Equilibriu and Elasticity.00 y y = ( 3) 2 / c 0. Figure P2.0 shows three unifor objects: a rod, a right triangle, and a square. Their asses in kilogras and their coordinates in eters are given. Deterine the center of gravity for the three-object syste kg ( 5,5) Figure P2.8 y() (2,7) 6.00 kg (8,5) (9,7) 3.00 kg 5.0 Figure P2. 3. A 5.0- unifor ladder weighing 500 N rests against a frictionless wall. The ladder akes a 60.0 angle with the horizontal. (a) Find the horizontal and vertical forces that the ground eerts on the base of the ladder when an 800-N firefighter is 4.00 fro the botto. (b) If the ladder is just on the verge of slipping when the firefighter is 9.00 up, what is the coefficient of static friction between the ladder and the ground? 4. A unifor ladder of length L and ass rests against a frictionless wall. The ladder akes an angle with the horizontal. (a) Find the horizontal and vertical forces that the ground eerts on the base of the ladder when a firefighter of ass 2 is a distance fro the botto. (b) If the ladder is just on the verge of slipping when the firefighter is a distance d fro the botto, what is the coefficient of static friction between the ladder and the ground? 5. Figure P2.5 shows a claw haer as it is being used to pull a nail out of a horizontal surface. If a force of agnitude 50 N is eerted horizontally as shown, find (a) the force eerted by the haer claws on the nail and (b) the force eerted by the surface on the point of contact with the haer head. Assue that the force the haer eerts on the nail is parallel to the nail. 6. A unifor plank with a length of 6.00 and a ass of 30.0 kg rests horizontally across two horizontal bars of a scaffold. The bars are 4.50 apart, and.50 of the plank hangs over one side of the scaffold. Draw a freebody diagra for the plank. How far can a painter with a ass of 70.0 kg walk on the overhanging part of the plank before it tips? 7. A 500-kg autoobile has a wheel base (the distance between the ales) of The center of ass of the autoobile is on the center line at a point.20 behind the front ale. Find the force eerted by the ground on each wheel. 8. A vertical post with a square cross section is 0.0 tall. Its botto end is encased in a base.50 tall that is precisely square but slightly loose. A force of 5.50 N to the right acts on the top of the post. The base aintains the post in equilibriu. Find the force that the top of the right sidewall of the base eerts on the post. Find the force that the botto of the left sidewall of the base eerts on the post. 9. A fleible chain weighing 40.0 N hangs between two hooks located at the sae height (Fig. P2.9). At each hook, the tangent to the chain akes an angle 42.0 with the horizontal. Find (a) the agnitude of the force each hook eerts on the chain and (b) the tension in the chain at its idpoint. (Hint: For part (b), ake a free-body diagra for half the chain.) 0.75 LuLu s Boutique Figure P ( 2,2) (4,) Figure P2.0 () Section 2.3 Eaples of Rigid bjects in Static Equilibriu. Stephen is pushing his sister Joyce in a wheelbarrow when it is stopped by a brick 8.00 c high (Fig. P2.). The handles ake an angle of 5.0 fro the horizontal. A downward force of 400 N is eerted on the wheel, which has a radius of 20.0 c. (a) What force ust Stephen apply along the handles to just start the wheel over the brick? (b) What is the force (agnitude and direction) that the brick eerts on the wheel just as the wheel begins to lift over the brick? Assue in both parts (a) and (b) that the brick reains fied and does not slide along the ground. 2. Two pans of a balance are 50.0 c apart. The fulcru of the balance has been shifted.00 c away fro the center by a dishonest shopkeeper. By what percentage is the true weight of the goods being arked up by the shopkeeper? (Assue that the balance has negligible ass.) c F 30.0 c Single point of contact Figure P2.5 θ Figure P A heispherical sign.00 in diaeter and of unifor ass density is supported by two strings, as shown in Figure P2.20. What fraction of the sign s weight is supported by each string? 2. Sir Lost-a-Lot dons his aror and sets out fro the castle on his trusty steed in his quest to iprove counication between dasels and dragons (Fig. P2.2). Unfortunately, his squire lowered the draw bridge too far and finally stopped lowering it when it was 20.0 below the horizontal. Lost-a-Lot and his horse stop when their cobined center of ass is.00 fro the end of the bridge. The bridge is 8.00 long and has a ass of kg. The lift cable is attached to the bridge 5.00 fro the hinge at the castle end and to a point on the castle wall 2.0 above the bridge. Lost-a-Lot s ass Figure P2.2 cobined with that of his aror and steed is 000 kg. Deterine (a) the tension in the cable, as well as (b) the horizontal and (c) the vertical force coponents acting on the bridge at the hinge. 22. Two identical, unifor bricks of length L are placed in a stack over the edge of a horizontal surface such that the aiu possible overhang without falling is achieved, as shown in Figure P2.22. Find the distance. Figure P2.22 L

69 Probles CHAPTER 2 Static Equilibriu and Elasticity A 23. A vaulter holds a 29.4-N pole in equilibriu by eerting an upward force U with her leading hand and a downward force D with her trailing hand, as shown in Figure P2.23. Point C is the center of gravity of the pole. What are the agnitudes of U and D? D B U Section C F g Figure P2.23 Elastic Properties of Solids Assue that Young s odulus for bone is N/ 2 and that a bone will fracture if ore than N/ 2 is eerted. (a) What is the aiu force that can be eerted on the feur bone in the leg if it has a iniu effective diaeter of 2.50 c? (b) If a force of this agnitude is applied copressively, by how uch does the 25.0-c-long bone shorten? 25. A 200-kg load is hung on a wire with a length of 4.00, a cross-sectional area of , and a Young s odulus of N/ 2. What is its increase in length? 26. A steel wire in diaeter can support a tension of 0.2 kn. Suppose you need a cable ade of these wires to support a tension of 20 kn. The cable s diaeter should be of what order of agnitude? 27. A child slides across a floor in a pair of rubber-soled shoes. The frictional force acting on each foot is 20.0 N. The footprint area of each shoe s sole is 4.0 c 2, and the thickness of each sole is Find the horizontal distance by which the upper and lower surfaces of each sole are offset. The shear odulus of the rubber is N/ Review Proble. A 30.0-kg haer strikes a steel spike 2.30 c in diaeter while oving with a speed of 20.0 /s. The haer rebounds with a speed of 0.0 /s after 0.0 s. What is the average strain in the spike during the ipact? 29. If the elastic liit of copper is N/ 2, deterine the iniu diaeter a copper wire can have under a load of 0.0 kg if its elastic liit is not to be eceeded. WEB 30. Review Proble. A long cylindrical steel wire with a cross-sectional diaeter of 4.00 is placed over a light frictionless pulley, with one end of the wire connected to a 5.00-kg ass and the other end connected to a 3.00-kg ass. By how uch does the wire stretch while the asses are in otion? 3. Review Proble. A cylindrical steel wire of length L i with a cross-sectional diaeter d is placed over a light frictionless pulley, with one end of the wire connected to a ass and the other end connected to a ass 2. By how uch does the wire stretch while the asses are in otion? 32. Calculate the density of sea water at a depth of 000, where the water pressure is about N/ 2. (The density of sea water is kg/ 3 at the surface.) 33. If the shear stress eceeds about N/ 2, steel ruptures. Deterine the shearing force necessary (a) to shear a steel bolt.00 c in diaeter and (b) to punch a.00-c-diaeter hole in a steel plate c thick. 34. (a) Find the iniu diaeter of a steel wire 8.0 long that elongates no ore than 9.00 when a load of 380 kg is hung on its lower end. (b) If the elastic liit for this steel is N/ 2, does peranent deforation occur with this load? 35. When water freezes, it epands by about 9.00%. What would be the pressure increase inside your autoobile s engine block if the water in it froze? (The bulk odulus of ice is N/ 2.) 36. For safety in clibing, a ountaineer uses a nylon rope that is 0.0 in diaeter. When supporting the 90.0-kg cliber on one end, the rope elongates by.60. Find Young s odulus for the rope aterial. ADDITINAL PRBLEMS 37. A bridge with a length of 50.0 and a ass of kg is supported on a sooth pier at each end, as illustrated in Figure P2.37. A truck of ass kg A Figure P2.37 B is located 5.0 fro one end. What are the forces on the bridge at the points of support? 38. A frae in the shape of the letter A is fored fro two unifor pieces of etal, each of weight 26.0 N and length.00. They are hinged at the top and held together by a horizontal wire.20 in length (Fig. P2.38). The structure rests on a frictionless surface. If the wire is connected at points a distance of fro the top of the frae, deterine the tension in the wire..20 Figure P Refer to Figure 2.7c. A lintel of prestressed reinforced concrete is.50 long. The cross-sectional area of the concrete is 50.0 c 2. The concrete encloses one steel reinforcing rod with a cross-sectional area of.50 c 2. The rod joins two strong end plates. Young s odulus for the concrete is N/ 2. After the concrete cures and the original tension T in the rod is released, the concrete will be under a copressive stress of N/ 2. (a) By what distance will the rod copress the concrete when the original tension in the rod is released? (b) Under what tension T 2 will the rod still be? (c) How uch longer than its unstressed length will the rod then be? (d) When the concrete was poured, the rod should have been stretched by what etension distance fro its unstressed length? (e) Find the required original tension T in the rod. 40. A solid sphere of radius R and ass M is placed in a trough, as shown in Figure P2.40. The inner surfaces of the trough are frictionless. Deterine the forces eerted by the trough on the sphere at the two contact points. 4. A 0.0-kg onkey clibs up a 20-N unifor ladder of length L, as shown in Figure P2.4. The upper and α 0.65 Figure P2.40 β 0.35 Figure P2.4 lower ends of the ladder rest on frictionless surfaces. The lower end is fastened to the wall by a horizontal rope that can support a aiu tension of 0 N. (a) Draw a free-body diagra for the ladder. (b) Find the tension in the rope when the onkey is one third the way up the ladder. (c) Find the aiu distance d that the onkey can clib up the ladder before the rope breaks. Epress your answer as a fraction of L. 42. A hungry bear weighing 700 N walks out on a bea in an attept to retrieve a basket of food hanging at the end of the bea (Fig. P2.42). The bea is unifor, weighs 200 N, and is 6.00 long; the basket weighs 80.0 N. (a) Draw a free-body diagra for the bea. (b) When the bear is at.00, find the tension in the wire and the coponents of the force eerted by the wall on the left end of the bea. (c) If the wire can withstand a aiu tension of 900 N, what is the aiu distance that the bear can walk before the wire breaks? 60.0 Goodies Figure P ld MacDonald had a far, and on that far he had a gate (Fig. P2.43). The gate is 3.00 wide and L

70 Probles CHAPTER 2 Static Equilibriu and Elasticity d 2L Figure P2.45 θ N where T is the force eerted by the Achilles tendon on the foot and R is the force eerted by the tibia on the foot. Find the values of T, R, and when F g 700 N. 5. A person bends over and lifts a 200-N object as shown in Figure P2.5a, with his back in a horizontal position (a terrible way to lift an object). The back uscle attached at a point two thirds the way up the spine aintains the position of the back, and the angle between the spine and this uscle is 2.0. Using the echanical odel shown in Figure P2.5b and taking the weight of the upper body to be 350 N, find the tension in the back uscle and the copressional force in the spine. WEB 3.00 Figure P2.43 high, with hinges attached to the top and botto. The guy wire akes an angle of 30.0 with the top of the gate and is tightened by a turn buckle to a tension of 200 N. The ass of the gate is 40.0 kg. (a) Deterine the horizontal force eerted on the gate by the botto hinge. (b) Find the horizontal force eerted by the upper hinge. (c) Deterine the cobined vertical force eerted by both hinges. (d) What ust the tension in the guy wire be so that the horizontal force eerted by the upper hinge is zero? 44. A 200-N unifor boo is supported by a cable, as illustrated in Figure P2.44. The boo is pivoted at the botto, and a N object hangs fro its top. Find the tension in the cable and the coponents of the reaction force eerted on the boo by the floor N 45. A unifor sign of weight F g and width 2L hangs fro a light, horizontal bea hinged at the wall and supported by a cable (Fig. P2.45). Deterine (a) the tension in the cable and (b) the coponents of the reaction force eerted by the wall on the bea in ters of F g, d, L, and. Figure P A crane of ass kg supports a load of kg as illustrated in Figure P2.46. The crane is pivoted with a frictionless pin at A and rests against a sooth support at B. Find the reaction forces at A and B..00 A (3 000 kg)g B Figure P kg 47. A ladder having a unifor density and a ass rests against a frictionless vertical wall, aking an angle 60.0 with the horizontal. The lower end rests on a flat surface, where the coefficient of static friction is s A window cleaner having a ass M 2 attepts to clib the ladder. What fraction of the length L of the ladder will the worker have reached when the ladder begins to slip? 48. A unifor ladder weighing 200 N is leaning against a wall (see Fig. 2.0). The ladder slips when Assuing that the coefficients of static friction at the wall and the ground are the sae, obtain a value for s. 49. A N shark is supported by a cable attached to a rod that can pivot at its base. Calculate the tension in the tie-rope between the wall and the rod if it is holding the syste in the position shown in Figure P2.49. Find the horizontal and vertical forces eerted on the base of the rod. (Neglect the weight of the rod.) 60.0 Figure P When a person stands on tiptoe (a strenuous position), the position of the foot is as shown in Figure P2.50a. The total weight of the body F g is supported by the force n eerted by the floor on the toe. A echanical odel for the situation is shown in Figure P2.50b, n Tibia (a) 5.0 R 8.0 c 25.0 c (b) θ Figure P2.50 Achilles tendon T 90.0 Pivot Back uscle (a) Figure P Two 200-N traffic lights are suspended fro a single cable, as shown in Figure Neglecting the cable s weight, (a) prove that if 2, then T T 2. (b) Deterine the three tensions T, T 2, and T 3 if T T 2 θ θ T 2 3 Figure P A force acts on a rectangular cabinet weighing 400 N, as illustrated in Figure P2.53. (a) If the cabinet slides with constant speed when F 200 N and h 0.400, R y R T N (b) 200 N

71 Probles CHAPTER 2 Static Equilibriu and Elasticity find the coefficient of kinetic friction and the position of the resultant noral force. (b) If F 300 N, find the value of h for which the cabinet just begins to tip. be suspended fro the top before the bea slips. (b) Deterine the agnitude of the reaction force at the floor and the agnitude of the force eerted by the bea on the rope at P in ters of, M, and s. C P w = 60 c 2.00 = 00 c 37.0 F P h M Figure P2.62 Figure P2.53 Probles 53 and Consider the rectangular cabinet of Proble 53, but with a force F applied horizontally at its upper edge. (a) What is the iniu force that ust be applied for the cabinet to start tipping? (b) What is the iniu coefficient of static friction required to prevent the cabinet fro sliding with the application of a force of this agnitude? (c) Find the agnitude and direction of the iniu force required to tip the cabinet if the point of application can be chosen anywhere on it. 55. A unifor rod of weight F g and length L is supported at its ends by a frictionless trough, as shown in Figure P2.55. (a) Show that the center of gravity of the rod is directly over point when the rod is in equilibriu. (b) Deterine the equilibriu value of the angle. θ Figure P2.58 shows a truss that supports a downward force of 000 N applied at the point B. The truss has negligible weight. The piers at A and C are sooth. (a) Apply the conditions of equilibriu to prove that n A 366 N and that n C 634 N. (b) Show that, because forces act on the light truss only at the hinge joints, each bar of the truss ust eert on each hinge pin only a force along the length of that bar a force of tension or copression. (c) Find the force of tension or copression in each of the three bars. A n A 0.0 θ Figure P N B n C C A 60. A flat dance floor of diensions 20.0 by 20.0 has a ass of 000 kg. Three dance couples, each of ass 25 kg, start in the top left, top right, and botto left corners. (a) Where is the initial center of gravity? (b) The couple in the botto left corner oves 0.0 to the right. Where is the new center of gravity? (c) What was the average velocity of the center of gravity if it took that couple 8.00 s to change position? 6. A shelf bracket is ounted on a vertical wall by a single screw, as shown in Figure P2.6. Neglecting the weight of the bracket, find the horizontal coponent of the force that the screw eerts on the bracket when an 80.0-N vertical force is applied as shown. (Hint: Iagine that the bracket is slightly loose.) 80.0 N 5.00 c 2.00 Figure P2.59 B static friction between the cylinder and all surfaces is In ters of F g, find the aiu force P that can be applied that does not cause the cylinder to rotate. (Hint: When the cylinder is on the verge of slipping, both friction forces are at their aiu values. Why?) WEB 63. Review Proble. A wire of length L i, Young s odulus Y, and cross-sectional area A is stretched elastically by an aount L. According to Hooke s law, the restoring force is k L. (a) Show that k YA/L i. (b) Show that the work done in stretching the wire by an aount L is W YA(L) 2 /2L i. 64. Two racquetballs are placed in a glass jar, as shown in Figure P2.64. Their centers and the point A lie on a straight line. (a) Assuing that the walls are frictionless, deterine P, P 2, and P 3. (b) Deterine the agnitude of the force eerted on the right ball by the left ball. Assue each ball has a ass of 70 g. Figure P2.55 Figure P c P 56. Review Proble. A cue stick strikes a cue ball and delivers a horizontal ipulse in such a way that the ball rolls without slipping as it starts to ove. At what height above the ball s center (in ters of the radius of the ball) was the blow struck? 57. A unifor bea of ass is inclined at an angle to the horizontal. Its upper end produces a 90 bend in a very rough rope tied to a wall, and its lower end rests on a rough floor (Fig. P2.57). (a) If the coefficient of static friction between the bea and the floor is s, deterine an epression for the aiu ass M that can 59. A stepladder of negligible weight is constructed as shown in Figure P2.59. A painter with a ass of 70.0 kg stands on the ladder 3.00 fro the botto. Assuing that the floor is frictionless, find (a) the tension in the horizontal bar connecting the two halves of the ladder, (b) the noral forces at A and B, and (c) the coponents of the reaction force at the single hinge C that the left half of the ladder eerts on the right half. (Hint: Treat each half of the ladder separately.) Figure P c 62. Figure P2.62 shows a vertical force applied tangentially to a unifor cylinder of weight F g. The coefficient of A P 3 P 2 Figure P In Figure P2.65, the scales read F g 380 N and F g N. Neglecting the weight of the supporting plank,

72 Probles CHAPTER 2 Static Equilibriu and Elasticity 2.00 F g F g 2 Figure P2.65 how far fro the woan s feet is her center of ass, given that her height is 2.00? 66. A steel cable 3.00 c 2 in cross-sectional area has a ass of 2.40 kg per eter of length. If 500 of the cable is hung over a vertical cliff, how uch does the cable stretch under its own weight? (For Young s odulus for steel, refer to Table 2..) 67. (a) Estiate the force with which a karate aster strikes a board if the hand s speed at tie of ipact is 0.0 /s and decreases to.00 /s during a s tie-of-contact with the board. The ass of coordinated hand-and-ar is.00 kg. (b) Estiate the shear stress if this force is eerted on a.00-c-thick pine board that is 0.0 c wide. (c) If the aiu shear stress a pine board can receive before breaking is N/ 2, will the board break? 68. A bucket is ade fro thin sheet etal. The botto and top of the bucket have radii of 25.0 c and 35.0 c, respectively. The bucket is 30.0 c high and filled with water. Where is the center of gravity? (Ignore the weight of the bucket itself.) 69. Review Proble. A trailer with a loaded weight of F g is being pulled by a vehicle with a force P, as illustrated in Figure P2.69. The trailer is loaded such that its center of ass is located as shown. Neglect the force of rolling friction and let a represent the coponent of the acceleration of the trailer. (a) Find the vertical coponent of P in ters of the given paraeters. (b) If a 2.00 /s 2 and h.50, what ust be the value of d so that P y 0 (that is, no vertical load on the vehicle)? (c) Find the values of P and P y given that F g 500 N, d 0.800, L 3.00, h.50, and a 2.00 /s Review Proble. An aluinu wire is long and has a circular cross section of diaeter Fied at the top end, the wire supports a.20-kg ass that swings in a horizontal circle. Deterine the angular velocity required to produce strain A 200--long bridge truss etends across a river (Fig. P2.7). Calculate the force of tension or copression in each structural coponent when a 360-kg car is at the center of the bridge. Assue that the structure is free to slide horizontally to perit theral epansion and contraction, that the structural coponents are connected by pin joints, and that the asses of the structural coponents are sall copared with the ass of the car. A B C 200 Figure P A 00--long bridge truss is supported at its ends so that it can slide freely (Fig. P2.72). A 500-kg car is halfway between points A and C. Show that the weight of the car is evenly distributed between points A and C, and calculate the force in each structural coponent. Specify whether each structural coponent is under tension or copression. Assue that the structural coponents are connected by pin joints and that the asses of the coponents are sall copared with the ass of the car. D E ANSWERS T QUICK QUIZZES 2. (a) Yes, as Figure 2.3 shows. The unbalanced torques cause an angular acceleration even though the linear acceleration is zero. (b) Yes, again. This happens when the lines of action of all the forces intersect at a coon point. If a net force acts on the object, then the object has a translational acceleration. However, because there is no net torque on the object, the object has no angular acceleration. There are other instances in which torques cancel but the forces do not. You should be able to draw at least two. 2.2 The location of the board s center of gravity relative to the fulcru. 2.3 Young s odulus is given by the ratio of stress to strain, which is the slope of the elastic behavior section of the graph in Figure 2.4. Reading fro the graph, we note that a stress of approiately N/ 2 results in a strain of The slope, and hence Young s odulus, are therefore N/ A substantial part of the graph etends beyond the elastic liit, indicating peranent deforation. Thus, the aterial is ductile. L B D d CM h P A C E n F g Figure P Figure P2.72

73 P U Z Z L E R Inside the pocket watch is a sall disk (called a torsional pendulu) that oscillates back and forth at a very precise rate and controls the watch gears. A grandfather clock keeps accurate tie because of its pendulu. The tall wooden case provides the space needed by the long pendulu as it advances the clock gears with each swing. In both of these tiepieces, the vibration of a carefully shaped coponent is critical to accurate operation. What properties of oscillating objects ake the so useful in tiing devices? (Photograph of pocket watch, George Seple; photograph of grandfather clock, Charles D. Winters) 390 CHAPTER 3 scillatory Motion Avery special kind of otion occurs when the force acting on a body is proportional to the displaceent of the body fro soe equilibriu position. If this force is always directed toward the equilibriu position, repetitive backand-forth otion occurs about this position. Such otion is called periodic otion, haronic otion, oscillation, or vibration (the four ters are copletely equivalent). You are ost likely failiar with several eaples of periodic otion, such as the oscillations of a block attached to a spring, the swinging of a child on a playground swing, the otion of a pendulu, and the vibrations of a stringed usical instruent. In addition to these everyday eaples, nuerous other systes ehibit periodic otion. For eaple, the olecules in a solid oscillate about their equilibriu positions; electroagnetic waves, such as light waves, radar, and radio waves, are characterized by oscillating electric and agnetic field vectors; and in alternating-current electrical circuits, voltage, current, and electrical charge vary periodically with tie. Most of the aterial in this chapter deals with siple haronic otion, in which an object oscillates such that its position is specified by a sinusoidal function of tie with no loss in echanical energy. In real echanical systes, daping (frictional) forces are often present. These forces are considered in optional Section 3.6 at the end of this chapter. scillatory Motion Chapter utline 3. Siple Haronic Motion 3.2 The Block Spring Syste Revisited 3.3 Energy of the Siple Haronic scillator 3.4 The Pendulu 3.5 Coparing Siple Haronic Motion with Unifor Circular Motion 3.6 (ptional) Daped scillations 3.7 (ptional) Forced scillations c h a p t e r (a) (b) (c) F s = 0 = 0 = 0 F s F s = 0 Figure 3. A block attached to a spring oving on a frictionless surface. (a) When the block is displaced to the right of equilibriu ( 0), the force eerted by the spring acts to the left. (b) When the block is at its equilibriu position ( 0), the force eerted by the spring is zero. (c) When the block is displaced to the left of equilibriu ( 0), the force eerted by the spring acts to the right SIMPLE HARMNIC MTIN Consider a physical syste that consists of a block of ass attached to the end of a spring, with the block free to ove on a horizontal, frictionless surface (Fig. 3.). When the spring is neither stretched nor copressed, the block is at the position 0, called the equilibriu position of the syste. We know fro eperience that such a syste oscillates back and forth if disturbed fro its equilibriu position. We can understand the otion in Figure 3. qualitatively by first recalling that when the block is displaced a sall distance fro equilibriu, the spring eerts on the block a force that is proportional to the displaceent and given by Hooke s law (see Section 7.3): F s k (3.) We call this a restoring force because it is is always directed toward the equilibriu position and therefore opposite the displaceent. That is, when the block is displaced to the right of 0 in Figure 3., then the displaceent is positive and the restoring force is directed to the left. When the block is displaced to the left of 0, then the displaceent is negative and the restoring force is directed to the right. Applying Newton s second law to the otion of the block, together with Equation 3., we obtain F s k a a k (3.2) That is, the acceleration is proportional to the displaceent of the block, and its direction is opposite the direction of the displaceent. Systes that behave in this way are said to ehibit siple haronic otion. An object oves with siple haronic otion whenever its acceleration is proportional to its displaceent fro soe equilibriu position and is oppositely directed.

74 3. Siple Haronic Motion CHAPTER 3 scillatory Motion The inverse of the period is called the frequency f of the otion. The frequency represents the nuber of oscillations that the particle akes per unit tie: Frequency f T 2 (3.5) 8.2 & 8.3 Motion of paper An eperiental arrangeent that ehibits siple haronic otion is illustrated in Figure 3.2. A ass oscillating vertically on a spring has a pen attached to it. While the ass is oscillating, a sheet of paper is oved perpendicular to the direction of otion of the spring, and the pen traces out a wavelike pattern. In general, a particle oving along the ais ehibits siple haronic otion when, the particle s displaceent fro equilibriu, varies in tie according to the relationship (3.3) where A,, and are constants. To give physical significance to these constants, we have labeled a plot of as a function of t in Figure 3.3a. This is just the pattern that is observed with the eperiental apparatus shown in Figure 3.2. The aplitude A of the otion is the aiu displaceent of the particle in either the positive or negative direction. The constant is called the angular frequency of the otion and has units of radians per second. (We shall discuss the geoetric significance of in Section 3.2.) The constant angle, called the phase constant (or phase angle), is deterined by the initial displaceent and velocity of the particle. If the particle is at its aiu position A at t 0, then and the curve of versus t is as shown in Figure 3.3b. If the particle is at soe other position at t 0, the constants and A tell us what the position was at tie t 0. The quantity (t ) is called the phase of the otion and is useful in coparing the otions of two oscillators. Note fro Equation 3.3 that the trigonoetric function is periodic and repeats itself every tie t increases by 2 rad. The period T of the otion is the tie it takes for the particle to go through one full cycle. We say that the particle has ade one oscillation. This definition of T tells us that the value of at tie t equals the value of at tie t T. We can show that T 2/ by using the preceding observation that the phase (t ) increases by 2 rad in a tie T: Hence, T 2, or Figure 3.2 An eperiental apparatus for deonstrating siple haronic otion. A pen attached to the oscillating ass traces out a wavelike pattern on the oving chart paper. A cos(t ) t 2 (t T ) T 2 0 (3.4) Displaceent versus tie for siple haronic otion φ/ω φω A A A A (a) (b) Figure 3.3 (a) An t curve for a particle undergoing siple haronic otion. The aplitude of the otion is A, the period is T, and the phase constant is. (b) The t curve in the special case in which A at t 0 and hence 0. T t t Angular frequency Velocity in siple haronic otion Acceleration in siple haronic otion Maiu values of speed and acceleration in siple haronic otion The units of f are cycles per second s, or hertz (Hz). Rearranging Equation 3.5, we obtain the angular frequency: Quick Quiz 3. (3.6) What would the phase constant have to be in Equation 3.3 if we were describing an oscillating object that happened to be at the origin at t 0? Quick Quiz 3.2 An object undergoes siple haronic otion of aplitude A. Through what total distance does the object ove during one coplete cycle of its otion? (a) A/2. (b) A. (c) 2A. (d) 4A. We can obtain the linear velocity of a particle undergoing siple haronic otion by differentiating Equation 3.3 with respect to tie: The acceleration of the particle is 2f 2 v d A sin(t ) dt a dv 2 A cos(t ) dt (3.7) (3.8) Because A cos(t ), we can epress Equation 3.8 in the for a 2 (3.9) Fro Equation 3.7 we see that, because the sine function oscillates between, the etree values of v are A. Because the cosine function also oscillates between, Equation 3.8 tells us that the etree values of a are 2 A. Therefore, the aiu speed and the agnitude of the aiu acceleration of a particle oving in siple haronic otion are v a A (3.0) a a 2 A (3.) Figure 3.4a represents the displaceent versus tie for an arbitrary value of the phase constant. The velocity and acceleration curves are illustrated in Figure 3.4b and c. These curves show that the phase of the velocity differs fro the phase of the displaceent by /2 rad, or 90. That is, when is a aiu or a iniu, the velocity is zero. Likewise, when is zero, the speed is a aiu. T

75 3. Siple Haronic Motion CHAPTER 3 scillatory Motion i (a) v v i T A v a = ωa ω t t Properties of siple haronic otion The following properties of a particle oving in siple haronic otion are iportant: The acceleration of the particle is proportional to the displaceent but is in the opposite direction. This is the necessary and sufficient condition for siple haronic otion, as opposed to all other kinds of vibration. The displaceent fro the equilibriu position, velocity, and acceleration all vary sinusoidally with tie but are not in phase, as shown in Figure 3.4. The frequency and the period of the otion are independent of the aplitude. (We show this eplicitly in the net section.) (b) (c) a a a = ω 2 A Furtherore, note that the phase of the acceleration differs fro the phase of the displaceent by rad, or 80. That is, when is a aiu, a is a aiu in the opposite direction. The phase constant is iportant when we copare the otion of two or ore oscillating objects. Iagine two identical pendulu bobs swinging side by side in siple haronic otion, with one having been released later than the other. The pendulu bobs have different phase constants. Let us show how the phase constant and the aplitude of any particle oving in siple haronic otion can be deterined if we know the particle s initial speed and position and the angular frequency of its otion. Suppose that at t 0 the initial position of a single oscillator is i and its initial speed is v v i. Under these conditions, Equations 3.3 and 3.7 give i A cos (3.2) v i A sin (3.3) Dividing Equation 3.3 by Equation 3.2 eliinates A, giving v i / i tan, or tan v i (3.4) i Furtherore, if we square Equations 3.2 and 3.3, divide the velocity equation by 2, and then add ters, we obtain 2 i v i Using the identity sin 2 cos 2, we can solve for A: t 2 A 2 cos 2 A 2 sin 2 A 2 i v i 2 Figure 3.4 Graphical representation of siple haronic otion. (a) Displaceent versus tie. (b) Velocity versus tie. (c) Acceleration versus tie. Note that at any specified tie the velocity is 90 out of phase with the displaceent and the acceleration is 80 out of phase with the displaceent. (3.5) EXAMPLE 3. Quick Quiz 3.3 Can we use Equations 2.8, 2.0, 2., and 2.2 (see pages 35 and 36) to describe the otion of a siple haronic oscillator? An scillating bject An object oscillates with siple haronic otion along the ais. Its displaceent fro the origin varies with tie according to the equation where t is in seconds and the angles in the parentheses are in radians. (a) Deterine the aplitude, frequency, and period of the otion. By coparing this equation with Equation 3.3, the general equation for siple haronic otion A cos(t ) we see that A 4.00 and rad/s. Therefore, f /2 / Hz and T /f 2.00 s. (b) Calculate the velocity and acceleration of the object at any tie t. v d (4.00 ) sin dt t a dv dt (4.00 /s) sin t (c) Using the results of part (b), deterine the position, velocity, and acceleration of the object at t.00 s. 4 ( /s 2 ) cos t (4.00 ) cos t 4 4 d dt (t) (4.00 /s) cos t 4 4 d dt (t) Noting that the angles in the trigonoetric functions are in radians, we obtain, at t.00 s, (4.00 ) cos (4.00 )(0.707) v (4.00 /s) sin 5 4 (4.00 /s)(0.707) a ( /s 2 ) cos 5 4 ( /s 2 )(0.707) (d) Deterine the aiu speed and aiu acceleration of the object /s In the general epressions for v and a found in part (b), we use the fact that the aiu values of the sine and cosine functions are unity. Therefore, v varies between 4.00 /s, and a varies between /s 2. Thus, v a 4.00 /s 2.6 /s a /s /s 2 a We obtain the sae results using v and a a 2 a A A, where A 4.00 and rad/s. (e) Find the displaceent of the object between t 0 and t.00 s. 4 (4.00 ) cos /s 2 4

76 3.2 The Block Spring Syste Revisited CHAPTER 3 scillatory Motion The coordinate at t 0 is i (4.00 ) cos 0 4 (4.00 )(0.707) 2.83 In part (c), we found that the coordinate at t.00 s is 2.83 ; therefore, the displaceent between t 0 and t.00 s is f i Because the object s velocity changes sign during the first second, the agnitude of is not the sae as the distance traveled in the first second. (By the tie the first second is over, the object has been through the point 2.83 once, traveled to 4.00, and coe back to 2.83.) Eercise What is the phase of the otion at t 2.00 s? Answer 9/4 rad. Period and frequency for a block spring syste position and in the opposite direction (F k), the particle oves in siple haronic otion. Recall that the period of any siple haronic oscillator is T 2/ (Eq. 3.4) and that the frequency is the inverse of the period. We know fro Equations 3.6 and 3.7 that, so we can epress the period and frequency of the block spring syste as k/ T 2 2 k (3.8) 3.2 THE BLCK SPRING SYSTEM REVISITED Let us return to the block spring syste (Fig. 3.5). Again we assue that the surface is frictionless; hence, when the block is displaced fro equilibriu, the only force acting on it is the restoring force of the spring. As we saw in Equation 3.2, when the block is displaced a distance fro equilibriu, it eperiences an acceleration a (k/). If the block is displaced a aiu distance A at soe initial tie and then released fro rest, its initial acceleration at that instant is ka/ (its etree negative value). When the block passes through the equilibriu position 0, its acceleration is zero. At this instant, its speed is a aiu. The block then continues to travel to the left of equilibriu and finally reaches A, at which tie its acceleration is ka/ (aiu positive) and its speed is again zero. Thus, we see that the block oscillates between the turning points A. Let us now describe the oscillating otion in a quantitative fashion. Recall that a dv/dt d 2 /dt 2, and so we can epress Equation 3.2 as d 2 (3.6) dt 2 k If we denote the ratio k/ with the sybol 2, this equation becoes d 2 (3.7) dt 2 2 Now we require a solution to Equation 3.7 that is, a function (t) that satisfies this second-order differential equation. Because Equations 3.7 and 3.9 are equivalent, each solution ust be that of siple haronic otion: A cos(t ) To see this eplicitly, assue that A cos(t ). Then (a) (b) (c) a = 0 = 0 = 0 a a = 0 Figure 3.5 A block of ass attached to a spring on a frictionless surface undergoes siple haronic otion. (a) When the block is displaced to the right of equilibriu, the displaceent is positive and the acceleration is negative. (b) At the equilibriu position, 0, the acceleration is zero and the speed is a aiu. (c) When the block is displaced to the left of equilibriu, the displaceent is negative and the acceleration is positive. QuickLab Hang an object fro a rubber band and start it oscillating. Measure T. Now tie four identical rubber bands together, end to end. How should k for this longer band copare with k for the single band? Again, tie the oscillations with the sae object. Can you verify Equation 3.9? f T 2 k (3.9) That is, the frequency and period depend only on the ass of the block and on the force constant of the spring. Furtherore, the frequency and period are independent of the aplitude of the otion. As we ight epect, the frequency is greater for a stiffer spring (the stiffer the spring, the greater the value of k) and decreases with increasing ass. Special Case. Let us consider a special case to better understand the physical significance of Equation 3.3, the defining epression for siple haronic otion. We shall use this equation to describe the otion of an oscillating block spring syste. Suppose we pull the block a distance A fro equilibriu and then release it fro rest at this stretched position, as shown in Figure 3.6. ur solution for ust obey the initial conditions that i A and v i 0 at t 0. It does if we choose 0, which gives A cos t as the solution. To check this solution, we note that it satisfies the condition that i A at t 0 because cos 0. Thus, we see that A and contain the inforation on initial conditions. Now let us investigate the behavior of the velocity and acceleration for this special case. Because A cos t, v d A sin t dt a dv 2 A cos t dt Fro the velocity epression we see that, because sin 0 0, v i 0 at t 0, as we require. The epression for the acceleration tells us that a 2 A at t 0. Physically, this negative acceleration akes sense because the force acting on the block is directed to the left when the displaceent is positive. In fact, at the etree po- d A d dt dt cos(t ) A sin(t ) d 2 dt 2 A d dt sin(t ) 2 A cos(t ) = 0 A t = 0 i = A v i = 0 = A cos ωt ω Coparing the epressions for and d 2 /dt 2, we see that d 2 /dt 2 2, and Equation 3.7 is satisfied. We conclude that whenever the force acting on a particle is linearly proportional to the displaceent fro soe equilibriu Figure 3.6 A block spring syste that starts fro rest at i A. In this case, 0 and thus A cos t.

77 3.2 The Block Spring Syste Revisited CHAPTER 3 scillatory Motion T 2 T = A cos ωt ω 3T 2 t speed at 0 and (2) the force and acceleration are zero at this position. The graphs of these functions versus tie in Figure 3.7 correspond to the origin at. Quick Quiz 3.4 v v = ωa ω sin ωtω What is the solution for if the block is initially oving to the left in Figure 3.8? a T 2 T 2 T T sition shown in Figure 3.6, F s ka (to the left) and the initial acceleration is 2 A ka/. Another approach to showing that A cos t is the correct solution involves using the relationship tan (Eq. 3.4). Because v i 0 at t 0, tan 0 and thus 0. (The tangent of also equals zero, but gives the wrong value for i.) Figure 3.7 is a plot of displaceent, velocity, and acceleration versus tie for this special case. Note that the acceleration reaches etree values of 2 A while the displaceent has etree values of A because the force is aial at those positions. Furtherore, the velocity has etree values of A, which both occur at 0. Hence, the quantitative solution agrees with our qualitative description of this syste. Special Case 2. Now suppose that the block is given an initial velocity v i to the right at the instant it is at the equilibriu position, so that i 0 and v v i at t 0 (Fig. 3.8). The epression for ust now satisfy these initial conditions. Because the block is oving in the positive direction at t 0 and because i 0 at t 0, the epression for ust have the for A sin t. Applying Equation 3.4 and the initial condition that i 0 at t 0, we find that tan and Hence, Equation 3.3 becoes A cos ( t /2), which can be written A sin t. Furtherore, fro Equation 3.5 we see that A v i / ; therefore, we can epress as 3T 2 3T 2 a = ωω 2 A cos ωtω v i / i /2. v i sin t The velocity and acceleration in this case are t t v d v dt i cos t Figure 3.7 Displaceent, velocity, and acceleration versus tie for a block spring syste like the one shown in Figure 3.6, undergoing siple haronic otion under the initial conditions that at t 0, i A and v i 0 (Special Case ). The origins at correspond to Special Case 2, the block spring syste under the initial conditions shown in Figure 3.8. a dv v i sin t dt These results are consistent with the facts that () the block always has a aiu i = 0 t = 0 v = v i = 0 = A sin ωt ω Figure 3.8 The block spring syste starts its otion at the equilibriu position at t 0. If its initial velocity is v i to the right, the block s coordinate varies as (v i /) sin t. v i EXAMPLE 3.2 Watch ut for Potholes! A car with a ass of 300 kg is constructed so that its frae is supported by four springs. Each spring has a force constant of N/. If two people riding in the car have a cobined ass of 60 kg, find the frequency of vibration of the car after it is driven over a pothole in the road. We assue that the ass is evenly distributed. Thus, each spring supports one fourth of the load. The total ass is 460 kg, and therefore each spring supports 365 kg. EXAMPLE 3.3 A Block Spring Syste A block with a ass of 200 g is connected to a light spring for which the force constant is 5.00 N/ and is free to oscillate on a horizontal, frictionless surface. The block is displaced 5.00 c fro equilibriu and released fro rest, as shown in Figure 3.6. (a) Find the period of its otion. Fro Equations 3.6 and 3.7, we know that the angular frequency of any block spring syste is 5.00 N/ rad/s kg k and the period is T rad/s (b) Deterine the aiu speed of the block. We use Equation 3.0: 2 v a A (5.00 rad/s)( ) s /s Hence, the frequency of vibration is, fro Equation 3.9, f Eercise How long does it take the car to eecute two coplete vibrations? Answer.70 s. 2 2 k N/ 365 kg.8 Hz (c) What is the aiu acceleration of the block? We use Equation 3.: a a 2 A (5.00 rad/s) 2 ( ) (d) Epress the displaceent, speed, and acceleration as functions of tie. This situation corresponds to Special Case, where our solution is A cos t. Using this epression and the results fro (a), (b), and (c), we find that A cos t v A sin t a 2 A cos t (0.050 ) cos 5.00t (0.250 /s) sin 5.00t (.25 /s 2 ) cos 5.00t.25 /s 2 ENERGY F THE SIMPLE HARMNIC SCILLATR Let us eaine the echanical energy of the block spring syste illustrated in Figure 3.6. Because the surface is frictionless, we epect the total echanical energy to be constant, as was shown in Chapter 8. We can use Equation 3.7 to e-

78 3.3 Energy of the Siple Haronic scillator CHAPTER 3 scillatory Motion press the kinetic energy as K 2 v A 2 sin 2 (t ) (3.20) The elastic potential energy stored in the spring for any elongation is given by 2 k2 (see Eq. 8.4). Using Equation 3.3, we obtain U 2 k2 2 ka2 cos 2 (t ) (3.2) We see that K and U are always positive quantities. Because 2 k/, we can epress the total echanical energy of the siple haronic oscillator as E K U 2 ka2 [sin 2 (t ) cos 2 (t )] cos 2, Fro the identity sin 2 we see that the quantity in square brackets is unity. Therefore, this equation reduces to E 2 ka2 (3.22) That is, the total echanical energy of a siple haronic oscillator is a constant of the otion and is proportional to the square of the aplitude. Note that U is sall when K is large, and vice versa, because the su ust be constant. In fact, the total echanical energy is equal to the aiu potential energy stored in the spring when A because v 0 at these points and thus there is no kinetic energy. At the equilibriu position, where U 0 because 0, the total energy, all in the for of kinetic energy, is again 2 ka2. That is, Kinetic energy of a siple haronic oscillator Potential energy of a siple haronic oscillator Total energy of a siple haronic oscillator Velocity as a function of position for a siple haronic oscillator θ a in Figure 3.9b. Energy is continuously being transfored between potential energy stored in the spring and kinetic energy of the block. Figure 3.0 illustrates the position, velocity, acceleration, kinetic energy, and potential energy of the block spring syste for one full period of the otion. Most of the ideas discussed so far are incorporated in this iportant figure. Study it carefully. Finally, we can use the principle of conservation of energy to obtain the velocity for an arbitrary displaceent by epressing the total energy at soe arbitrary position as E K U 2 v 2 2 v k2 2 ka2 k (3.23) (A2 2 ) A 2 2 When we check Equation 3.23 to see whether it agrees with known cases, we find that it substantiates the fact that the speed is a aiu at 0 and is zero at the turning points A. v a a a t v a K U 0 A 0 ωω 2 A 0 2 ka2 E 2 v 2 a 2 2 A 2 2 k (at 0) A2 2 ka2 Plots of the kinetic and potential energies versus tie appear in Figure 3.9a, where we have taken 0. As already entioned, both K and U are always positive, and at all ties their su is a constant equal to 2 ka2, the total energy of the syste. The variations of K and U with the displaceent of the block are plotted θ a a a T/4 0 ωaω ka2 T/2 A 0 ω 2 A 0 2 ka2 K, U U K φ = 0 U = k 2 2 K = v 2 2 K, U v a 3T/4 0 ωaω ka2 2 ka2 T 2 (a) T t A Figure 3.9 (a) Kinetic energy and potential energy versus tie for a siple haronic oscillator with 0. (b) Kinetic energy and potential energy versus displaceent for a siple haronic oscillator. In either plot, note that K U constant. (b) A θ a A 0 A a a T A 0 ωω 2 A 0 2 ka2 Figure 3.0 Siple haronic otion for a block spring syste and its relationship to the otion of a siple pendulu. The paraeters in the table refer to the block spring syste, assuing that A at t 0; thus, A cos t (see Special Case ).

79 3.3 Energy of the Siple Haronic scillator CHAPTER 3 scillatory Motion U Using the result of (b), we find that K 2 v 2 2 (0.500 kg)(0.4 /s) J Note that K U E. Eercise For what values of is the speed of the cube 0.00 /s? r U 2 k2 2 (20.0 N/)( ) J Answer 2.55 c. Figure 3. (a) If the atos in a olecule do not ove too far fro their equilibriu positions, a graph of potential energy versus separation distance between atos is siilar to the graph of potential energy versus position for a siple haronic oscillator. (b) Tiny springs approiate the forces holding atos together. You ay wonder why we are spending so uch tie studying siple haronic oscillators. We do so because they are good odels of a wide variety of physical phenoena. For eaple, recall the Lennard Jones potential discussed in Eaple 8.. This coplicated function describes the forces holding atos together. Figure 3.a shows that, for sall displaceents fro the equilibriu position, the potential energy curve for this function approiates a parabola, which represents the potential energy function for a siple haronic oscillator. Thus, we can approiate the cople atoic binding forces as tiny springs, as depicted in Figure 3.b. The ideas presented in this chapter apply not only to block spring systes and atos, but also to a wide range of situations that include bungee juping, tuning in a television station, and viewing the light eitted by a laser. You will see ore eaples of siple haronic oscillators as you work through this book. EXAMPLE 3.4 scillations on a Horizontal Surface A kg cube connected to a light spring for which the force constant is 20.0 N/ oscillates on a horizontal, frictionless track. (a) Calculate the total energy of the syste and the aiu speed of the cube if the aplitude of the otion is 3.00 c. Using Equation 3.22, we obtain E K U 2 ka2 2 (20.0 N/) ( ) J When the cube is at 0, we know that U 0 and E 2 v 2 a ; therefore, 2 v 2 a J v a J kg 0.90 /s (b) What is the velocity of the cube when the displaceent is 2.00 c? We can apply Equation 3.23 directly: v k (A2 2 ) 20.0 N/ kg [( )2 ( ) 2 ] 0.4 /s The positive and negative signs indicate that the cube could be oving to either the right or the left at this instant. (c) Copute the kinetic and potential energies of the syste when the displaceent is 2.00 c. L θ T 8. & 8.2 s g sin θ θ g cos θ g Figure 3.2 When is sall, a siple pendulu oscillates in siple haronic otion about the equilibriu position 0. The restoring force is g sin, the coponent of the gravitational force tangent to the arc. 3.4 THE PENDULUM The siple pendulu is another echanical syste that ehibits periodic otion. It consists of a particle-like bob of ass suspended by a light string of length L that is fied at the upper end, as shown in Figure 3.2. The otion occurs in the vertical plane and is driven by the force of gravity. We shall show that, provided the angle is sall (less than about 0 ), the otion is that of a siple haronic oscillator. The forces acting on the bob are the force T eerted by the string and the gravitational force g. The tangential coponent of the gravitational force, g sin, always acts toward 0, opposite the displaceent. Therefore, the tangential force is a restoring force, and we can apply Newton s second law for otion in the tangential direction: F t g sin d 2 s dt 2 where s is the bob s displaceent easured along the arc and the inus sign indicates that the tangential force acts toward the equilibriu (vertical) position. Because s L (Eq. 0.a) and L is constant, this equation reduces to The right side is proportional to sin rather than to ; hence, with sin present, we would not epect siple haronic otion because this epression is not of the for of Equation 3.7. However, if we assue that is sall, we can use the approiation sin ; thus the equation of otion for the siple pend 2 dt 2 g L sin The otion of a siple pendulu, captured with ultiflash photography. Is the oscillating otion siple haronic in this case?

80 3.4 The Pendulu CHAPTER 3 scillatory Motion The period of the otion is dulu becoes d 2 (3.24) dt 2 g L Now we have an epression of the sae for as Equation 3.7, and we conclude that the otion for sall aplitudes of oscillation is siple haronic otion. Therefore, can be written as a cos(t ), where a is the aiu angular displaceent and the angular frequency is g L (3.25) Equation of otion for a siple pendulu (sall ) Angular frequency of otion for a siple pendulu Period of otion for a siple pendulu T 2 2 L g (3.26) In other words, the period and frequency of a siple pendulu depend only on the length of the string and the acceleration due to gravity. Because the period is independent of the ass, we conclude that all siple pendulus that are of equal length and are at the sae location (so that g is constant) oscillate with the sae period. The analogy between the otion of a siple pendulu and that of a block spring syste is illustrated in Figure 3.0. The siple pendulu can be used as a tiekeeper because its period depends only on its length and the local value of g. It is also a convenient device for aking precise easureents of the free-fall acceleration. Such easureents are iportant because variations in local values of g can provide inforation on the location of oil and of other valuable underground resources. Quick Quiz 3.5 A block of ass is first allowed to hang fro a spring in static equilibriu. It stretches the spring a distance L beyond the spring s unstressed length. The block and spring are then set into oscillation. Is the period of this syste less than, equal to, or greater than the period of a siple pendulu having a length L and a bob ass? EXAMPLE 3.5 A Connection Between Length and Tie Christian Huygens ( ), the greatest clockaker in history, suggested that an international unit of length could be defined as the length of a siple pendulu having a period of eactly s. How uch shorter would our length unit be had his suggestion been followed? L T 2 g Solving Equation 3.26 for the length gives 4 2 ( 4 s)2 (9.80 /s 2 ) Thus, the eter s length would be slightly less than onefourth its current length. Note that the nuber of significant digits depends only on how precisely we know g because the tie has been defined to be eactly s. The Foucault pendulu at the Franklin Institute in Philadelphia. This type of pendulu was first used by the French physicist Jean Foucault to verify the Earth s rotation eperientally. As the pendulu swings, the vertical plane in which it oscillates appears to rotate as the bob successively knocks over the indicators arranged in a circle on the floor. In reality, the plane of oscillation is fied in space, and the Earth rotating beneath the swinging pendulu oves the indicators into position to be knocked down, one after the other. QuickLab Firly hold a ruler so that about half of it is over the edge of your desk. With your other hand, pull down and then release the free end, watching how it vibrates. Now slide the ruler so that only about a quarter of it is free to vibrate. This tie when you release it, how does the vibrational period copare with its earlier value? Why? Physical Pendulu Suppose you balance a wire coat hanger so that the hook is supported by your etended inde finger. When you give the hanger a sall displaceent (with your other hand) and then release it, it oscillates. If a hanging object oscillates about a fied ais that does not pass through its center of ass and the object cannot be approiated as a point ass, we cannot treat the syste as a siple pendulu. In this case the syste is called a physical pendulu. Consider a rigid body pivoted at a point that is a distance d fro the center of ass (Fig. 3.3). The force of gravity provides a torque about an ais through, and the agnitude of that torque is gd sin, where is as shown in Figure 3.3. Using the law of otion where I is the oent of inertia about I,

81 3.4 The Pendulu CHAPTER 3 scillatory Motion the ais through, we obtain The inus sign indicates that the torque about tends to decrease. That is, the force of gravity produces a restoring torque. Because this equation gives us the angular acceleration d 2 /dt 2 of the pivoted body, we can consider it the equation of otion for the syste. If we again assue that is sall, the approiation sin is valid, and the equation of otion reduces to (3.27) Because this equation is of the sae for as Equation 3.7, the otion is siple haronic otion. That is, the solution of Equation 3.27 is a cos(t ), where a is the aiu angular displaceent and The period is gd sin I d 2 dt 2 d 2 dt 2 gd I 2 T 2 gd 2 I gd (3.28) ne can use this result to easure the oent of inertia of a flat rigid body. If the location of the center of ass and hence the value of d are known, the oent of inertia can be obtained by easuring the period. Finally, note that Equation 3.28 reduces to the period of a siple pendulu (Eq. 3.26) when I d 2 that is, when all the ass is concentrated at the center of ass. EXAMPLE 3.6 A Swinging Rod A unifor rod of ass M and length L is pivoted about one end and oscillates in a vertical plane (Fig. 3.4). Find the period of oscillation if the aplitude of the otion is sall. In Chapter 0 we found that the oent of inertia of a unifor rod about an ais through one end is 3 ML 2. The distance d fro the pivot to the center of ass is L/2. Substituting these quantities into Equation 3.28 gives T 3 2 3g ML 2 Mg L 2 2 2L Coent In one of the Moon landings, an astronaut walking on the Moon s surface had a belt hanging fro his space suit, and the belt oscillated as a physical pendulu. A scientist on the Earth observed this otion on television and used it to estiate the free-fall acceleration on the Moon. How did the scientist ake this calculation? I Pivot θ Period of otion for a physical pendulu d d sin θ g CM Figure 3.3 A physical pendulu. Eercise Calculate the period of a eter stick that is pivoted about one end and is oscillating in a vertical plane. Answer.64 s. Figure 3.4 L CM Mg Pivot A rigid rod oscillating about a pivot through one end is a physical pendulu with d L/2 and, fro Table 0.2, I 3 ML 2. θ a P Figure 3.5 A torsional pendulu consists of a rigid body suspended by a wire attached to a rigid support. The body oscillates about the line P with an aplitude a. Period of otion for a torsional pendulu 8.8 Torsional Pendulu Figure 3.5 shows a rigid body suspended by a wire attached at the top to a fied support. When the body is twisted through soe sall angle, the twisted wire eerts on the body a restoring torque that is proportional to the angular displaceent. That is, where (kappa) is called the torsion constant of the support wire. The value of can be obtained by applying a known torque to twist the wire through a easurable angle. Applying Newton s second law for rotational otion, we find dt 2 d 2 (3.29) dt 2 I Again, this is the equation of otion for a siple haronic oscillator, with and a period /I (3.30) This syste is called a torsional pendulu. There is no sall-angle restriction in this situation as long as the elastic liit of the wire is not eceeded. Figure 3.6 shows the balance wheel of a watch oscillating as a torsional pendulu, energized by the ainspring. 3.5 I d 2 T 2 I CMPARING SIMPLE HARMNIC MTIN WITH UNIFRM CIRCULAR MTIN We can better understand and visualize any aspects of siple haronic otion by studying its relationship to unifor circular otion. Figure 3.7 is an overhead view of an eperiental arrangeent that shows this relationship. A ball is attached to the ri of a turntable of radius A, which is illuinated fro the side by a lap. The ball casts a shadow on a screen. We find that as the turntable rotates with constant angular speed, the shadow of the ball oves back and forth in siple haronic otion. Balance wheel Figure 3.6 The balance wheel of this antique pocket watch is a torsional pendulu and regulates the tie-keeping echanis.

82 3.5 Coparing Siple Haronic Motion with Unifor Circular Motion CHAPTER 3 scillatory Motion Consider a particle located at point P on the circuference of a circle of radius A, as shown in Figure 3.8a, with the line P aking an angle with the ais at t 0. We call this circle a reference circle for coparing siple haronic otion and unifor circular otion, and we take the position of P at t 0 as our reference position. If the particle oves along the circle with constant angular speed until P akes an angle with the ais, as illustrated in Figure 3.8b, then at soe tie t 0, the angle between P and the ais is t. As the particle oves along the circle, the projection of P on the ais, labeled point Q, oves back and forth along the ais, between the liits A. Note that points P and Q always have the sae coordinate. Fro the right triangle PQ, we see that this coordinate is A cos(t ) (3.3) This epression shows that the point Q oves with siple haronic otion along the ais. Therefore, we conclude that siple haronic otion along a straight line can be represented by the projection of unifor circular otion along a diaeter of a reference circle. We can ake a siilar arguent by noting fro Figure 3.8b that the projection of P along the y ais also ehibits siple haronic otion. Therefore, unifor circular otion can be considered a cobination of two siple haronic otions, one along the ais and one along the y ais, with the two differing in phase by 90. This geoetric interpretation shows that the tie for one coplete revolution of the point P on the reference circle is equal to the period of otion T for siple haronic otion between A. That is, the angular speed of P is the sae as the angular frequency of siple haronic otion along the ais (this is why we use the sae sybol). The phase constant for siple haronic otion corresponds to the initial angle that P akes with the ais. The radius A of the reference circle equals the aplitude of the siple haronic otion. y (a) A φ P y y (b) A θ ω P Q t = 0 θ = ωt ω + φ v = ωa ω y v Figure 3.8 Relationship between the unifor circular otion of a point P and the siple haronic otion of a point Q. A particle at P oves in a circle of radius A with constant angular speed. (a) A reference circle showing the position of P at t 0. (b) The coordinates of points P and Q are equal and vary in tie as A cos(t ). (c) The coponent of the velocity of P equals the velocity of Q. (d) The coponent of the acceleration of P equals the acceleration of Q. (c) v v P Q a = ω 2 A y (d) a a Q a A A Q Lap P P Ball Turntable Screen Shadow of ball Figure 3.7 An eperiental setup for deonstrating the connection between siple haronic otion and unifor circular otion. As the ball rotates on the turntable with constant angular speed, its shadow on the screen oves back and forth in siple haronic otion. EXAMPLE 3.7 Because the relationship between linear and angular speed for circular otion is v r (see Eq. 0.0), the particle oving on the reference circle of radius A has a velocity of agnitude A. Fro the geoetry in Figure 3.8c, we see that the coponent of this velocity is A sin(t ). By definition, the point Q has a velocity given by d/dt. Differentiating Equation 3.3 with respect to tie, we find that the velocity of Q is the sae as the coponent of the velocity of P. The acceleration of P on the reference circle is directed radially inward toward and has a agnitude v 2 /A 2 A. Fro the geoetry in Figure 3.8d, we see that the coponent of this acceleration is 2 A cos(t ). This value is also the acceleration of the projected point Q along the ais, as you can verify by taking the second derivative of Equation 3.3. Circular Motion with Constant Angular Speed A particle rotates counterclockwise in a circle of radius 3.00 with a constant angular speed of 8.00 rad/s. At t 0, the particle has an coordinate of 2.00 and is oving to the right. (a) Deterine the coordinate as a function of tie. Because the aplitude of the particle s otion equals the radius of the circle and 8.00 rad/s, we have A cos(t ) (3.00 ) cos(8.00t ) We can evaluate by using the initial condition that 2.00 at t 0: 2.00 (3.00 ) cos(0 ) cos If we were to take our answer as 48.2, then the coordinate (3.00 ) cos (8.00t 48.2 ) would be decreasing at tie t 0 (that is, oving to the left). Because our particle is first oving to the right, we ust choose rad. The coordinate as a function of tie is then ptional Section 3.6 DAMPED Note that in the cosine function ust be in radians. (b) Find the coponents of the particle s velocity and acceleration at any tie t. v d (3.00 )(8.00 rad/s) sin(8.00t 0.84) dt a dv dt (3.00 ) cos (8.00t 0.84) (24.0 /s) sin(8.00t 0.84) (24.0 /s)(8.00 rad/s) cos(8.00t 0.84) (92 /s 2 ) cos(8.00t 0.84) Fro these results, we conclude that v a 24.0 /s and that a a 92 /s 2. Note that these values also equal the tangential speed A and the centripetal acceleration 2 A. SCILLATINS The oscillatory otions we have considered so far have been for ideal systes that is, systes that oscillate indefinitely under the action of a linear restoring force. In any real systes, dissipative forces, such as friction, retard the otion. Consequently, the echanical energy of the syste diinishes in tie, and the otion is said to be daped. ne coon type of retarding force is the one discussed in Section 6.4, where the force is proportional to the speed of the oving object and acts in the direction opposite the otion. This retarding force is often observed when an object oves through air, for instance. Because the retarding force can be epressed as R bv (where b is a constant called the daping coefficient) and the restoring

83 3.6 Daped scillations CHAPTER 3 scillatory Motion A b t 2 Ae 0 t (a) il or other viscous fluid Piston with holes Coil spring Shock absorber force of the syste is k, we can write Newton s second law as F k bv a k b d d 2 (3.32) dt dt 2 The solution of this equation requires atheatics that ay not be failiar to you yet; we siply state it here without proof. When the retarding force is sall copared with the aiu restoring force that is, when b is sall the solution to Equation 3.32 is Ae b 2 t cos(t ) (3.33) where the angular frequency of oscillation is (3.34) 2 b 2 This result can be verified by substituting Equation 3.33 into Equation Figure 3.9a shows the displaceent as a function of tie for an object oscillating in the presence of a retarding force, and Figure 3.9b depicts one such syste: a block attached to a spring and subersed in a viscous liquid. We see that when the retarding force is uch saller than the restoring force, the oscillatory character of the otion is preserved but the aplitude decreases in tie, with the result that the otion ultiately ceases. Any syste that behaves in this way is known as a daped oscillator. The dashed blue lines in Figure 3.9a, which define the envelope of the oscillatory curve, represent the eponential factor in Equation This envelope shows that the aplitude decays eponentially with tie. For otion with a given spring constant and block ass, the oscillations dapen ore rapidly as the aiu value of the retarding force approaches the aiu value of the restoring force. It is convenient to epress the angular frequency of a daped oscillator in the for 2 b 2 2 k 0 where 0 k/ represents the angular frequency in the absence of a retarding force (the undaped oscillator) and is called the natural frequency of the syste. When the agnitude of the aiu retarding force R a bv a ka, the syste is said to be underdaped. As the value of R approaches ka, the aplitudes of the oscillations decrease ore and ore rapidly. This otion is represented by the blue curve in Figure When b reaches a critical value b c such that b c /2 0, the syste does not oscillate and is said to be critically daped. In this case the syste, once released fro rest at soe nonequilibriu position, returns to equilibriu and then stays there. The graph of displaceent versus tie for this case is the red curve in Figure If the ediu is so viscous that the retarding force is greater than the restoring force that is, if R a bv a ka and b/2 0 the syste is overdaped. Again, the displaced syste, when free to ove, does not oscillate but siply returns to its equilibriu position. As the daping increases, the tie it takes the syste to approach equilibriu also increases, as indicated by the black curve in Figure In any case in which friction is present, whether the syste is overdaped or underdaped, the energy of the oscillator eventually falls to zero. The lost echanical energy dissipates into internal energy in the retarding ediu. (b) Figure 3.9 (a) Graph of displaceent versus tie for a daped oscillator. Note the decrease in aplitude with tie. (b) ne eaple of a daped oscillator is a ass attached to a spring and subersed in a viscous liquid. a b Figure 3.20 Graphs of displaceent versus tie for (a) an underdaped oscillator, (b) a critically daped oscillator, and (c) an overdaped oscillator. c t web To learn ore about shock absorbers, visit Quick Quiz 3.6 An autootive suspension syste consists of a cobination of springs and shock absorbers, as shown in Figure 3.2. If you were an autootive engineer, would you design a suspension syste that was underdaped, critically daped, or overdaped? Discuss each case. ptional Section 3.7 (a) Figure 3.2 (a) A shock absorber consists of a piston oscillating in a chaber filled with oil. As the piston oscillates, the oil is squeezed through holes between the piston and the chaber, causing a daping of the piston s oscillations. (b) ne type of autootive suspension syste, in which a shock absorber is placed inside a coil spring at each wheel. FRCEDSCILLATINS It is possible to copensate for energy loss in a daped syste by applying an eternal force that does positive work on the syste. At any instant, energy can be put into the syste by an applied force that acts in the direction of otion of the oscillator. For eaple, a child on a swing can be kept in otion by appropriately tied pushes. The aplitude of otion reains constant if the energy input per cycle eactly equals the energy lost as a result of daping. Any otion of this type is called forced oscillation. A coon eaple of a forced oscillator is a daped oscillator driven by an eternal force that varies periodically, such as F F et cos t, where is the angular frequency of the periodic force and F et is a constant. Adding this driving force to the left side of Equation 3.32 gives F et cos t k b d d 2 (3.35) dt dt 2 (As earlier, we present the solution of this equation without proof.) After a sufficiently long period of tie, when the energy input per cycle equals the energy lost per cycle, a steady-state condition is reached in which the oscillations proceed with constant aplitude. At this tie, when the syste is in a steady state, the solution of Equation 3.35 is A cos(t ) (3.36) (b)

84 3.7 Forced scillations 4 42 CHAPTER 3 scillatory Motion where A F et / (2 0 2 ) 2 b 2 (3.37) and where 0 k/ is the angular frequency of the undaped oscillator (b 0). ne could argue that in steady state the oscillator ust physically have the sae frequency as the driving force, and thus the solution given by Equation 3.36 is epected. In fact, when this solution is substituted into Equation 3.35, one finds that it is indeed a solution, provided the aplitude is given by Equation Equation 3.37 shows that, because an eternal force is driving it, the otion of the forced oscillator is not daped. The eternal agent provides the necessary energy to overcoe the losses due to the retarding force. Note that the syste oscillates at the angular frequency of the driving force. For sall daping, the aplitude becoes very large when the frequency of the driving force is near the natural frequency of oscillation. The draatic increase in aplitude near the natural frequency 0 is called resonance, and for this reason 0 is soeties called the resonance frequency of the syste. The reason for large-aplitude oscillations at the resonance frequency is that energy is being transferred to the syste under the ost favorable conditions. We can better understand this by taking the first tie derivative of in Equation 3.36, which gives an epression for the velocity of the oscillator. We find that v is proportional to sin(t ). When the applied force F is in phase with the velocity, the rate at which work is done on the oscillator by F equals the dot product F v. Reeber that rate at which work is done is the definition of power. Because the product F v is a aiu when F and v are in phase, we conclude that at resonance the applied force is in phase with the velocity and that the power transferred to the oscillator is a aiu. Figure 3.22 is a graph of aplitude as a function of frequency for a forced oscillator with and without daping. Note that the aplitude increases with decreasing daping (b : 0) and that the resonance curve broadens as the daping increases. Under steady-state conditions and at any driving frequency, the energy transferred into the syste equals the energy lost because of the daping force; hence, the average total energy of the oscillator reains constant. In the absence of a daping force (b 0), we see fro Equation 3.37 that the steady-state aplitude approaches infinity as : 0. In other words, if there are no losses in the syste and if we continue to drive an initially otionless oscillator with a periodic force that is in phase with the velocity, the aplitude of otion builds without liit (see the red curve in Fig. 3.22). This liitless building does not occur in practice because soe daping is always present. The behavior of a driven oscillating syste after the driving force is reoved depends on b and on how close was to 0. This behavior is soeties quantified by a paraeter called the quality factor Q. The closer a syste is to being undaped, the greater its Q. The aplitude of oscillation drops by a factor of e ( ) in Q/ cycles. Later in this book we shall see that resonance appears in other areas of physics. For eaple, certain electrical circuits have natural frequencies. A bridge has natural frequencies that can be set into resonance by an appropriate driving force. A draatic eaple of such resonance occurred in 940, when the Tacoa Narrows Bridge in the state of Washington was destroyed by resonant vibrations. Although the winds were not particularly strong on that occasion, the bridge ultiately collapsed (Fig. 3.23) because the bridge design had no built-in safety features. 0 A ω 0 b = 0 Undaped Sall b Large b Figure 3.22 Graph of aplitude versus frequency for a daped oscillator when a periodic driving force is present. When the frequency of the driving force equals the natural frequency 0, resonance occurs. Note that the shape of the resonance curve depends on the size of the daping coefficient b. QuickLab Tie several objects to strings and suspend the fro a horizontal string, as illustrated in the figure. Make two of the hanging strings approiately the sae length. If one of this pair, such as P, is set into sideways otion, all the others begin to oscillate. But Q, whose length is the sae as that of P, oscillates with the greatest aplitude. Must all the asses have the sae value? P Q ω (a) Figure 3.23 (a) In 940 turbulent winds set up torsional vibrations in the Tacoa Narrows Bridge, causing it to oscillate at a frequency near one of the natural frequencies of the bridge structure. (b) nce established, this resonance condition led to the bridge s collapse. Many other eaples of resonant vibrations can be cited. A resonant vibration that you ay have eperienced is the singing of telephone wires in the wind. Machines often break if one vibrating part is at resonance with soe other oving part. Soldiers arching in cadence across a bridge have been known to set up resonant vibrations in the structure and thereby cause it to collapse. Whenever any real physical syste is driven near its resonance frequency, you can epect oscillations of very large aplitudes. SUMMARY When the acceleration of an object is proportional to its displaceent fro soe equilibriu position and is in the direction opposite the displaceent, the object oves with siple haronic otion. The position of a siple haronic oscillator varies periodically in tie according to the epression A cos(t ) (3.3) where A is the aplitude of the otion, is the angular frequency, and is the phase constant. The value of depends on the initial position and initial velocity of the oscillator. You should be able to use this forula to describe the otion of an object undergoing siple haronic otion. The tie T needed for one coplete oscillation is defined as the period of the otion: T 2 (3.4) The inverse of the period is the frequency of the otion, which equals the nuber of oscillations per second. The velocity and acceleration of a siple haronic oscillator are v d A sin(t ) dt a dv 2 A cos(t ) dt v A 2 2 (b) (3.7) (3.8) (3.23)

85 Questions CHAPTER 3 scillatory Motion Thus, the aiu speed is A, and the aiu acceleration is 2 A. The speed is zero when the oscillator is at its turning points, A, and is a aiu when the oscillator is at the equilibriu position 0. The agnitude of the acceleration is a aiu at the turning points and zero at the equilibriu position. You should be able to find the velocity and acceleration of an oscillating object at any tie if you know the aplitude, angular frequency, and phase constant. A block spring syste oves in siple haronic otion on a frictionless surface, with a period T 2 2 (3.8) k The kinetic energy and potential energy for a siple haronic oscillator vary with tie and are given by K 2 v A 2 sin 2 (t ) (3.20) U 2 k2 2 ka2 cos 2 (t ) (3.2) These three forulas allow you to analyze a wide variety of situations involving oscillations. Be sure you recognize how the ass of the block and the spring constant of the spring enter into the calculations. The total energy of a siple haronic oscillator is a constant of the otion and is given by E 2 ka2 (3.22) The potential energy of the oscillator is a aiu when the oscillator is at its turning points and is zero when the oscillator is at the equilibriu position. The kinetic energy is zero at the turning points and a aiu at the equilibriu position. You should be able to deterine the division of energy between potential and kinetic fors at any tie t. A siple pendulu of length L oves in siple haronic otion. For sall angular displaceents fro the vertical, its period is L T 2 (3.26) For sall angular displaceents fro g the vertical, a physical pendulu oves in siple haronic otion about a pivot that does not go through the center of ass. The period of this otion is I T 2 (3.28) gd where I is the oent of inertia about an ais through the pivot and d is the distance fro the pivot to the center of ass. You should be able to distinguish when to use the siple-pendulu forula and when the syste ust be considered a physical pendulu. Unifor circular otion can be considered a cobination of two siple haronic otions, one along the ais and the other along the y ais, with the two differing in phase by 90. QUESTINS. Is a bouncing ball an eaple of siple haronic otion? Is the daily oveent of a student fro hoe to school and back siple haronic otion? Why or why not? 2. If the coordinate of a particle varies as Acos t, what is the phase constant in Equation 3.3? At what position does the particle begin its otion? 3. Does the displaceent of an oscillating particle between t 0 and a later tie t necessarily equal the position of the particle at tie t? Eplain. 4. Deterine whether the following quantities can be in the sae direction for a siple haronic oscillator: (a) displaceent and velocity, (b) velocity and acceleration, (c) displaceent and acceleration. 5. Can the aplitude A and the phase constant be deterined for an oscillator if only the position is specified at t 0? Eplain. 6. Describe qualitatively the otion of a ass spring syste when the ass of the spring is not neglected. 7. Make a graph showing the potential energy of a stationary block hanging fro a spring, U 2 ky2 gy. Why is the lowest part of the graph offset fro the origin? 8. A block spring syste undergoes siple haronic otion with an aplitude A. Does the total energy change if the ass is doubled but the aplitude is not changed? Do the kinetic and potential energies depend on the ass? Eplain. 9. What happens to the period of a siple pendulu if the pendulu s length is doubled? What happens to the period if the ass of the suspended bob is doubled? 0. A siple pendulu is suspended fro the ceiling of a stationary elevator, and the period is deterined. Describe the changes, if any, in the period when the elevator PRBLEMS, 2, 3 = straightforward, interediate, challenging = full solution available in the Student s Manual and Study Guide WEB = solution posted at = Coputer useful in solving proble = Interactive Physics = paired nuerical/sybolic probles Section 3. Siple Haronic Motion. The displaceent of a particle at t s is given by the epression (4.00 ) cos(3.00t ), where is in eters and t is in seconds. Deterine (a) the frequency and period of the otion, (b) the aplitude of the otion, (c) the phase constant, and (d) the displaceent of the particle at t s. 2. A ball dropped fro a height of 4.00 akes a perfectly elastic collision with the ground. Assuing that no energy is lost due to air resistance, (a) show that the otion is periodic and (b) deterine the period of the otion. (c) Is the otion siple haronic? Eplain. 3. A particle oves in siple haronic otion with a frequency of 3.00 oscillations/s and an aplitude of 5.00 c. (a) Through what total distance does the particle ove during one cycle of its otion? (b) What is its aiu speed? Where does this occur? (c) Find the aiu acceleration of the particle. Where in the otion does the aiu acceleration occur? 4. In an engine, a piston oscillates with siple haronic otion so that its displaceent varies according to the epression (5.00 c) cos(2t /6) where is in centieters and t is in seconds. At t 0, (a) accelerates upward, (b) accelerates downward, and (c) oves with constant velocity.. A siple pendulu undergoes siple haronic otion when is sall. Is the otion periodic when is large? How does the period of otion change as increases? 2. Will daped oscillations occur for any values of b and k? Eplain. 3. As it possible to have daped oscillations when a syste is at resonance? Eplain. 4. At resonance, what does the phase constant equal in Equation 3.36? (Hint: Copare this equation with the epression for the driving force, which ust be in phase with the velocity at resonance.) 5. Soe parachutes have holes in the to allow air to ove soothly through the. Without such holes, soeties the air that has gathered beneath the chute as a parachutist falls is released fro under its edges alternately and periodically, at one side and then at the other. Why ight this periodic release of air cause a proble? 6. If a grandfather clock were running slowly, how could we adjust the length of the pendulu to correct the tie? 7. A pendulu bob is ade fro a sphere filled with water. What would happen to the frequency of vibration of this pendulu if the sphere had a hole in it that allowed the water to leak out slowly? WEB find (a) the displaceent of the particle, (b) its velocity, and (c) its acceleration. (d) Find the period and aplitude of the otion. 5. A particle oving along the ais in siple haronic otion starts fro its equilibriu position, the origin, at t 0 and oves to the right. The aplitude of its otion is 2.00 c, and the frequency is.50 Hz. (a) Show that the displaceent of the particle is given by (2.00 c) sin(3.00t). Deterine (b) the aiu speed and the earliest tie (t 0) at which the particle has this speed, (c) the aiu acceleration and the earliest tie (t 0) at which the particle has this acceleration, and (d) the total distance traveled between t 0 and t.00 s. 6. The initial position and initial velocity of an object oving in siple haronic otion are i and v i ; the angular frequency of oscillation is. (a) Show that the position and velocity of the object for all tie can be written as (t) i cos t v i sin t v(t) i sin t v i cos t (b) If the aplitude of the otion is A, show that v 2 a v 2 i a i i 2 A 2

86 Probles CHAPTER 3 scillatory Motion Section 3.2 The Block Spring Syste Revisited Note: Neglect the ass of the spring in all probles in this section. 7. A spring stretches by 3.90 c when a 0.0-g ass is hung fro it. If a 25.0-g ass attached to this spring oscillates in siple haronic otion, calculate the period of the otion. 8. A siple haronic oscillator takes 2.0 s to undergo five coplete vibrations. Find (a) the period of its otion, (b) the frequency in hertz, and (c) the angular frequency in radians per second. 9. A kg ass attached to a spring with a force constant of 8.00 N/ vibrates in siple haronic otion with an aplitude of 0.0 c. Calculate (a) the aiu value of its speed and acceleration, (b) the speed and acceleration when the ass is 6.00 c fro the equilibriu position, and (c) the tie it takes the ass to ove fro 0 to 8.00 c. 0. A.00-kg ass attached to a spring with a force constant of 25.0 N/ oscillates on a horizontal, frictionless track. At t 0, the ass is released fro rest at 3.00 c. (That is, the spring is copressed by 3.00 c.) Find (a) the period of its otion; (b) the aiu values of its speed and acceleration; and (c) the displaceent, velocity, and acceleration as functions of tie.. A 7.00-kg ass is hung fro the botto end of a vertical spring fastened to an overhead bea. The ass is set into vertical oscillations with a period of 2.60 s. Find the force constant of the spring. 2. A block of unknown ass is attached to a spring with a spring constant of 6.50 N/ and undergoes siple haronic otion with an aplitude of 0.0 c. When the ass is halfway between its equilibriu position and the end point, its speed is easured to be 30.0 c/s. Calculate (a) the ass of the block, (b) the period of the otion, and (c) the aiu acceleration of the block. 3. A particle that hangs fro a spring oscillates with an angular frequency of 2.00 rad/s. The spring particle syste is suspended fro the ceiling of an elevator car and hangs otionless (relative to the elevator car) as the car descends at a constant speed of.50 /s. The car then stops suddenly. (a) With what aplitude does the particle oscillate? (b) What is the equation of otion for the particle? (Choose upward as the positive direction.) 4. A particle that hangs fro a spring oscillates with an angular frequency. The spring particle syste is suspended fro the ceiling of an elevator car and hangs otionless (relative to the elevator car) as the car descends at a constant speed v. The car then stops suddenly. (a) With what aplitude does the particle oscillate? (b) What is the equation of otion for the particle? (Choose upward as the positive direction.) 5. A.00-kg ass is attached to a horizontal spring. The spring is initially stretched by 0.00, and the ass is released fro rest there. It proceeds to ove without friction. After s, the speed of the ass is zero. What is the aiu speed of the ass? Section 3.3 Energy of the Siple Haronic scillator Note: Neglect the ass of the spring in all probles in this section. 6. A 200-g ass is attached to a spring and undergoes siple haronic otion with a period of s. If the total energy of the syste is 2.00 J, find (a) the force constant of the spring and (b) the aplitude of the otion. 7. An autoobile having a ass of 000 kg is driven into a brick wall in a safety test. The buper behaves as a spring of constant N/ and copresses 3.6 c as the car is brought to rest. What was the speed of the car before ipact, assuing that no energy is lost during ipact with the wall? 8. A ass spring syste oscillates with an aplitude of 3.50 c. If the spring constant is 250 N/ and the ass is kg, deterine (a) the echanical energy of the syste, (b) the aiu speed of the ass, and (c) the aiu acceleration. 9. A 50.0-g ass connected to a spring with a force constant of 35.0 N/ oscillates on a horizontal, frictionless surface with an aplitude of 4.00 c. Find (a) the total energy of the syste and (b) the speed of the ass when the displaceent is.00 c. Find (c) the kinetic energy and (d) the potential energy when the displaceent is 3.00 c. 20. A 2.00-kg ass is attached to a spring and placed on a horizontal, sooth surface. A horizontal force of 20.0 N is required to hold the ass at rest when it is pulled fro its equilibriu position (the origin of the ais). The ass is now released fro rest with an initial displaceent of i 0.200, and it subsequently undergoes siple haronic oscillations. Find (a) the force constant of the spring, (b) the frequency of the oscillations, and (c) the aiu speed of the ass. Where does this aiu speed occur? (d) Find the aiu acceleration of the ass. Where does it occur? (e) Find the total energy of the oscillating syste. Find (f) the speed and (g) the acceleration when the displaceent equals one third of the aiu value. 2. A.50-kg block at rest on a tabletop is attached to a horizontal spring having force constant of 9.6 N/. The spring is initially unstretched. A constant 20.0-N horizontal force is applied to the object, causing the spring to stretch. (a) Deterine the speed of the block after it has oved fro equilibriu, assuing that the surface between the block and the tabletop is frictionless. (b) Answer part (a) for a coefficient of kinetic friction of between the block and the tabletop. 22. The aplitude of a syste oving in siple haronic otion is doubled. Deterine the change in (a) the total energy, (b) the aiu speed, (c) the aiu acceleration, and (d) the period. WEB 23. A particle eecutes siple haronic otion with an aplitude of 3.00 c. At what displaceent fro the idpoint of its otion does its speed equal one half of its aiu speed? 24. A ass on a spring with a constant of 3.24 N/ vibrates, with its position given by the equation (5.00 c) cos(3.60t rad/s). (a) During the first cycle, for 0 t.75 s, when is the potential energy of the syste changing ost rapidly into kinetic energy? (b) What is the aiu rate of energy transforation? Section 3.4 The Pendulu 25. A an enters a tall tower, needing to know its height. He notes that a long pendulu etends fro the ceiling alost to the floor and that its period is 2.0 s. (a) How tall is the tower? (b) If this pendulu is taken to the Moon, where the free-fall acceleration is.67 /s 2, what is its period there? 26. A seconds pendulu is one that oves through its equilibriu position once each second. (The period of the pendulu is s.) The length of a seconds pendulu is at Tokyo and at Cabridge, England. What is the ratio of the free-fall accelerations at these two locations? 27. A rigid steel frae above a street intersection supports standard traffic lights, each of which is hinged to hang iediately below the frae. A gust of wind sets a light swinging in a vertical plane. Find the order of agnitude of its period. State the quantities you take as data and their values. 28. The angular displaceent of a pendulu is represented by the equation (0.320 rad)cos t, where is in radians and 4.43 rad/s. Deterine the period and length of the pendulu. 29. A siple pendulu has a ass of kg and a length of.00. It is displaced through an angle of 5.0 and then released. What are (a) the aiu speed, (b) the aiu angular acceleration, and (c) the aiu restoring force? 30. A siple pendulu is 5.00 long. (a) What is the period of siple haronic otion for this pendulu if it is hanging in an elevator that is accelerating upward at 5.00 /s 2? (b) What is its period if the elevator is accelerating downward at 5.00 /s 2? (c) What is the period of siple haronic otion for this pendulu if it is placed in a truck that is accelerating horizontally at 5.00 /s 2? 3. A particle of ass slides without friction inside a heispherical bowl of radius R. Show that, if it starts fro rest with a sall displaceent fro equilibriu, the particle oves in siple haronic otion with an angular frequency equal to that of a siple pendulu of length R. That is, 32. A ass is attached to the end of a string to for a siple pendulu. The period of its haronic otion is WEB g/r. easured for sall angular displaceents and three lengths; in each case, the otion is clocked with a stopwatch for 50 oscillations. For lengths of.000, 0.750, and 0.500, total ties of 99.8 s, 86.6 s, and 7. s, respectively, are easured for the 50 oscillations. (a) Deterine the period of otion for each length. (b) Deterine the ean value of g obtained fro these three independent easureents, and copare it with the accepted value. (c) Plot T 2 versus L, and obtain a value for g fro the slope of your best-fit straight-line graph. Copare this value with that obtained in part (b). 33. A physical pendulu in the for of a planar body oves in siple haronic otion with a frequency of Hz. If the pendulu has a ass of 2.20 kg and the pivot is located fro the center of ass, deterine the oent of inertia of the pendulu. 34. A very light, rigid rod with a length of etends straight out fro one end of a eter stick. The stick is suspended fro a pivot at the far end of the rod and is set into oscillation. (a) Deterine the period of oscillation. (b) By what percentage does this differ fro a.00--long siple pendulu? 35. Consider the physical pendulu of Figure 3.3. (a) If I CM is its oent of inertia about an ais that passes through its center of ass and is parallel to the ais that passes through its pivot point, show that its period is T 2 I CM d 2 gd where d is the distance between the pivot point and the center of ass. (b) Show that the period has a iniu value when d satisfies d 2 I CM. 36. A torsional pendulu is fored by attaching a wire to the center of a eter stick with a ass of 2.00 kg. If the resulting period is 3.00 in, what is the torsion constant for the wire? 37. A clock balance wheel has a period of oscillation of s. The wheel is constructed so that 20.0 g of ass is concentrated around a ri of radius c. What are (a) the wheel s oent of inertia and (b) the torsion constant of the attached spring? Section 3.5 Coparing Siple Haronic Motion with Unifor Circular Motion 38. While riding behind a car that is traveling at 3.00 /s, you notice that one of the car s tires has a sall heispherical boss on its ri, as shown in Figure P3.38. (a) Eplain why the boss, fro your viewpoint behind the car, eecutes siple haronic otion. (b) If the radius of the car s tires is 0.300, what is the boss s period of oscillation? 39. Consider the siplified single-piston engine shown in Figure P3.39. If the wheel rotates with constant angular speed, eplain why the piston rod oscillates in siple haronic otion.

87 Probles CHAPTER 3 scillatory Motion ω A (ptional) Section 3.6 Boss Figure P3.38 Figure P3.39 Daped scillations Piston = A (t ) (ptional) Section 3.7 Forced scillations 43. A baby rejoices in the day by crowing and juping up and down in her crib. Her ass is 2.5 kg, and the crib attress can be odeled as a light spring with a force constant of 4.30 kn/. (a) The baby soon learns to bounce with aiu aplitude and iniu effort by bending her knees at what frequency? (b) She learns to use the attress as a trapoline losing contact with it for part of each cycle when her aplitude eceeds what value? 44. A 2.00-kg ass attached to a spring is driven by an eternal force F (3.00 N) cos(2t). If the force constant of the spring is 20.0 N/, deterine (a) the pe- 40. Show that the tie rate of change of echanical energy for a daped, undriven oscillator is given by de/dt bv 2 and hence is always negative. (Hint: Differentiate the epression for the echanical energy of an oscillator, E 2 v 2 2 k2, and use Eq ) 4. A pendulu with a length of.00 is released fro an initial angle of 5.0. After 000 s, its aplitude is reduced by friction to What is the value of b/2? 42. Show that Equation 3.33 is a solution of Equation 3.32 provided that b 2 4k. riod and (b) the aplitude of the otion. (Hint: Assue that there is no daping that is, that b 0 and use Eq ) 45. Considering an undaped, forced oscillator (b 0), show that Equation 3.36 is a solution of Equation 3.35, with an aplitude given by Equation A weight of 40.0 N is suspended fro a spring that has a force constant of 200 N/. The syste is undaped and is subjected to a haronic force with a frequency of 0.0 Hz, which results in a forced-otion aplitude of 2.00 c. Deterine the aiu value of the force. 47. Daping is negligible for a 0.50-kg ass hanging fro a light 6.30-N/ spring. The syste is driven by a force oscillating with an aplitude of.70 N. At what frequency will the force ake the ass vibrate with an aplitude of 0.440? 48. You are a research biologist. Before dining at a fine restaurant, you set your pager to vibrate instead of beep, and you place it in the side pocket of your suit coat. The ar of your chair presses the light cloth against your body at one spot. Fabric with a length of 8.2 c hangs freely below that spot, with the pager at the botto. A co-worker telephones you. The otion of the vibrating pager akes the hanging part of your coat swing back and forth with rearkably large aplitude. The waiter, aître d, wine steward, and nearby diners notice iediately and fall silent. Your daughter pipes up and says, Daddy, look! Your cockroaches ust have gotten out again! Find the frequency at which your pager vibrates. ADDITINAL PRBLEMS 49. A car with bad shock absorbers bounces up and down with a period of.50 s after hitting a bup. The car has a ass of 500 kg and is supported by four springs of equal force constant k. Deterine the value of k. 50. A large passenger with a ass of 50 kg sits in the iddle of the car described in Proble 49. What is the new period of oscillation? 5. A copact ass M is attached to the end of a unifor rod, of equal ass M and length L, that is pivoted at the top (Fig. P3.5). (a) Deterine the tensions in the rod L P M Pivot y Figure P3.5 y = 0 WEB at the pivot and at the point P when the syste is stationary. (b) Calculate the period of oscillation for sall displaceents fro equilibriu, and deterine this period for L (Hint: Assue that the ass at the end of the rod is a point ass, and use Eq ) 52. A ass, 9.00 kg, is in equilibriu while connected to a light spring of constant k 00 N/ that is fastened to a wall, as shown in Figure P3.52a. A second ass, kg, is slowly pushed up against ass, copressing the spring by the aount A (see Fig. P3.52b). The syste is then released, and both asses start oving to the right on the frictionless surface. (a) When reaches the equilibriu point, 2 loses contact with (see Fig. P3.52c) and oves to the right with speed v. Deterine the value of v. (b) How far apart are the asses when the spring is fully stretched for the first tie (D in Fig. P3.52d)? (Hint: First deterine the period of oscillation and the aplitude of the spring syste after 2 loses contact with.) (a) (b) (c) (d) k k k 2 k A v 2 Figure P3.52 v A large block P eecutes horizontal siple haronic otion as it slides across a frictionless surface with a frequency of f.50 Hz. Block B rests on it, as shown D µ s Figure P3.53 Probles 53 and 54. in Figure P3.53, and the coefficient of static friction between the two is s What aiu aplitude of oscillation can the syste have if block B is not to slip? 54. A large block P eecutes horizontal siple haronic otion as it slides across a frictionless surface with a frequency f. Block B rests on it, as shown in Figure P3.53, and the coefficient of static friction between the two is s. What aiu aplitude of oscillation can the syste have if the upper block is not to slip? 55. The ass of the deuteriu olecule (D 2 ) is twice that of the hydrogen olecule (H 2 ). If the vibrational frequency of H 2 is Hz, what is the vibrational frequency of D 2? Assue that the spring constant of attracting forces is the sae for the two olecules. 56. A solid sphere (radius R) rolls without slipping in a cylindrical trough (radius 5R), as shown in Figure P3.56. Show that, for sall displaceents fro equilibriu perpendicular to the length of the trough, the sphere eecutes siple haronic otion with a period T 2 28R/5g. 5R 57. A light cubical container of volue a 3 is initially filled with a liquid of ass density. The container is initially supported by a light string to for a pendulu of length L i, easured fro the center of ass of the filled container. The liquid is allowed to flow fro the botto of the container at a constant rate (dm/dt). At any tie t, the level of the liquid in the container is h B P R Figure P3.56

88 Probles CHAPTER 3 scillatory Motion and the length of the pendulu is L (easured relative to the instantaneous center of ass). (a) Sketch the apparatus and label the diensions a, h, L i, and L. (b) Find the tie rate of change of the period as a function of tie t. (c) Find the period as a function of tie. 58. After a thrilling plunge, bungee-jupers bounce freely on the bungee cords through any cycles. Your little brother can ake a pest of hiself by figuring out the ass of each person, using a proportion he set up by solving this proble: A ass is oscillating freely on a vertical spring with a period T (Fig. P3.58a). An unknown ass on the sae spring oscillates with a period T. Deterine (a) the spring constant k and (b) the unknown ass. (a) Figure P3.58 (a) Mass spring syste for Probles 58 and 68. (b) Bungee-juping fro a bridge. (Telegraph Colour Library/ FPG International) 59. A pendulu of length L and ass M has a spring of force constant k connected to it at a distance h below its point of suspension (Fig. P3.59). Find the frequency of vibration of the syste for sall values of the aplitude (sall ). (Assue that the vertical suspension of length L is rigid, but neglect its ass.) 60. A horizontal plank of ass and length L is pivoted at one end. The plank s other end is supported by a spring of force constant k (Fig. P3.60). The oent of inertia of the plank about the pivot is 3 L 2. (a) Show that the plank, after being displaced a sall angle fro its horizontal equilibriu position and released, oves with siple haronic otion of angular frequency (b) Evaluate the frequency if the ass is 5.00 kg and the spring has a force constant of 00 N/. 3k/. (b) h θ L Pivot M 6. ne end of a light spring with a force constant of 00 N/ is attached to a vertical wall. A light string is tied to the other end of the horizontal spring. The string changes fro horizontal to vertical as it passes over a 4.00-c-diaeter solid pulley that is free to turn on a fied sooth ale. The vertical section of the string supports a 200-g ass. The string does not slip at its contact with the pulley. Find the frequency of oscillation of the ass if the ass of the pulley is (a) negligible, (b) 250 g, and (c) 750 g. 62. A 2.00-kg block hangs without vibrating at the end of a spring (k 500 N/) that is attached to the ceiling of an elevator car. The car is rising with an upward acceleration of g/3 when the acceleration suddenly ceases (at t 0). (a) What is the angular frequency of oscillation of the block after the acceleration ceases? (b) By what aount is the spring stretched during the acceleration of the elevator car? (c) What are the aplitude of the oscillation and the initial phase angle observed by a rider in the car? Take the upward direction to be positive. 63. A siple pendulu with a length of 2.23 and a ass of 6.74 kg is given an initial speed of 2.06 /s at its equilibriu position. Assue that it undergoes siple haronic otion, and deterine its (a) period, (b) total energy, and (c) aiu angular displaceent. θ k Figure P3.59 Figure P3.60 k WEB 64. People who ride otorcycles and bicycles learn to look out for bups in the road and especially for washboarding, which is a condition of any equally spaced ridges worn into the road. What is so bad about washboarding? A otorcycle has several springs and shock absorbers in its suspension, but you can odel it as a single spring supporting a ass. You can estiate the spring constant by thinking about how far the spring copresses when a big biker sits down on the seat. A otorcyclist traveling at highway speed ust be particularly careful of washboard bups that are a certain distance apart. What is the order of agnitude of their separation distance? State the quantities you take as data and the values you estiate or easure for the. 65. A wire is bent into the shape of one cycle of a cosine curve. It is held in a vertical plane so that the height y of the wire at any horizontal distance fro the center is given by y 20.0 c[ cos(0.60 rad/)]. A bead can slide without friction on the stationary wire. Show that if its ecursion away fro 0 is never large, the bead oves with siple haronic otion. Deterine its angular frequency. (Hint: cos for sall.) 66. A block of ass M is connected to a spring of ass and oscillates in siple haronic otion on a horizontal, frictionless track (Fig. P3.66). The force constant of the spring is k, and the equilibriu length is. Find (a) the kinetic energy of the syste when the block has a speed v, and (b) the period of oscillation. (Hint: Assue that all portions of the spring oscillate in phase and that the velocity of a segent d is proportional to the distance fro the fied end; that is, v [/]v. Also, note that the ass of a segent of the spring is d [/]d.) 67. A ball of ass is connected to two rubber bands of length L, each under tension T, as in Figure P3.67. The ball is displaced by a sall distance y perpendicular to the length of the rubber bands. Assuing that the tension does not change, show that (a) the restoring force is (2T/L)y and (b) the syste ehibits siple haronic otion with an angular frequency 2T/L. d Figure P /2 M v L 68. When a ass M, connected to the end of a spring of ass s 7.40 g and force constant k, is set into siple haronic otion, the period of its otion is T 2 M ( s/3) k A two-part eperient is conducted with the use of various asses suspended vertically fro the spring, as shown in Figure P3.58a. (a) Static etensions of 7.0, 29.3, 35.3, 4.3, 47., and 49.3 c are easured for M values of 20.0, 40.0, 50.0, 60.0, 70.0, and 80.0 g, respectively. Construct a graph of Mg versus, and perfor a linear least-squares fit to the data. Fro the slope of your graph, deterine a value for k for this spring. (b) The syste is now set into siple haronic otion, and periods are easured with a stopwatch. With M 80.0 g, the total tie for 0 oscillations is easured to be 3.4 s. The eperient is repeated with M values of 70.0, 60.0, 50.0, 40.0, and 20.0 g, with corresponding ties for 0 oscillations of 2.52,.67, 0.67, 9.62, and 7.03 s. Copute the eperiental value for T for each of these easureents. Plot a graph of T 2 versus M, and deterine a value for k fro the slope of the linear least-squares fit through the data points. Copare this value of k with that obtained in part (a). (c) btain a value for s fro your graph, and copare it with the given value of 7.40 g. 69. A sall, thin disk of radius r and ass is attached rigidly to the face of a second thin disk of radius R and ass M, as shown in Figure P3.69. The center of the sall disk is located at the edge of the large disk. The large disk is ounted at its center on a frictionless ale. The assebly is rotated through a sall angle fro its equilibriu position and released. (a) Show that the v M θ y Figure P3.67 θ R Figure P3.69 L

89 Answers to Quick Quizzes CHAPTER 3 scillatory Motion speed of the center of the sall disk as it passes through the equilibriu position is (b) Show that the period of the otion is 70. Consider the daped oscillator illustrated in Figure 3.9. Assue that the ass is 375 g, the spring constant is 00 N/, and b 0.00 kg/s. (a) How long does it takes for the aplitude to drop to half its initial value? (b) How long does it take for the echanical energy to drop to half its initial value? (c) Show that, in general, the fractional rate at which the aplitude decreases in a daped haronic oscillator is one-half the fractional rate at which the echanical energy decreases. 7. A ass is connected to two springs of force constants k and k 2, as shown in Figure P3.7a and b. In each case, the ass oves on a frictionless table and is displaced fro equilibriu and then released. Show that in the two cases the ass ehibits siple haronic otion with periods (a) (b) Rg( cos ) v 2 (M/) (r/r) 2 2 /2 T 2 (M 2)R 2 r 2 2gR /2 T 2 (k k 2 ) k k 2 T 2 k k Consider a siple pendulu of length L.20 that is displaced fro the vertical by an angle a and then released. You are to predict the subsequent angular displaceents when a is sall and also when it is large. Set up and carry out a nuerical ethod to integrate k k 2 (a) k k 2 (b) Figure P3.7 the equation of otion for the siple pendulu: d 2 dt 2 g L sin Take the initial conditions to be a and d/dt 0 at t 0. n one trial choose a 5.00, and on another trial take a 00. In each case, find the displaceent as a function of tie. Using the sae values for a, copare your results for with those obtained fro a cos t. How does the period for the large value of a copare with that for the sall value of a? Note: Using the Euler ethod to solve this differential equation, you ay find that the aplitude tends to increase with tie. The fourth-order Runge Kutta ethod would be a better choice to solve the differential equation. However, if you choose t sall enough, the solution that you obtain using Euler s ethod can still be good. 3.6 If your goal is siply to stop the bounce fro an absorbed shock as rapidly as possible, you should critically dap the suspension. Unfortunately, the stiffness of this design akes for an uncofortable ride. If you underdap the suspension, the ride is ore cofortable but the car bounces. If you overdap the suspension, the wheel is displaced fro its equilibriu position longer than it should be. (For eaple, after hitting a bup, the spring stays copressed for a short tie and the wheel does not quickly drop back down into contact with the road after the wheel is past the bup a dangerous situation.) Because of all these considerations, autootive engineers usually design suspensions to be slightly underdaped. This allows the suspension to absorb a shock rapidly (iniizing the roughness of the ride) and then return to equilibriu after only one or two noticeable oscillations. ANSWERS T QUICK QUIZZES 3. Because A can never be zero, ust be any value that results in the cosine function s being zero at t 0. In other words, cos (0). This is true at /2, 3/2 or, ore generally, n/2, where n is any nonzero odd integer. If we want to restrict our choices of to values between 0 and 2, we need to know whether the object was oving to the right or to the left at t 0. If it was oving with a positive velocity, then 3/2. If v i 0, then / (d) 4A. Fro its aiu positive position to the equilibriu position, it travels a distance A, by definition of aplitude. It then goes an equal distance past the equilibriu position to its aiu negative position. It then repeats these two otions in the reverse direction to return to its original position and coplete one cycle. 3.3 No, because in siple haronic otion, the acceleration is not constant. 3.4 A sin t, where A v i /. 3.5 Fro Hooke s law, the spring constant ust be k g/l. If we substitute this value for k into Equation 3.8, we find that T 2 k 2 g/l 2 L g This is the sae as Equation 3.26, which gives the period of a siple pendulu. Thus, when an object stretches a vertically hung spring, the period of the syste is the sae as that of a siple pendulu having a length equal to the aount of static etension of the spring.

90 The Law of Gravity Chapter utline 2.2 This is the Nearest ne Head 423 P U Z Z L E R More than 300 years ago, Isaac Newton realized that the sae gravitational force that causes apples to fall to the Earth also holds the Moon in its orbit. In recent years, scientists have used the Hubble Space Telescope to collect evidence of the gravitational force acting even farther away, such as at this protoplanetary disk in the constellation Taurus. What properties of an object such as a protoplanet or the Moon deterine the strength of its gravitational attraction to another object? (Left, Larry West/FPG International; right, Courtesy of NASA) web For ore inforation about the Hubble, visit the Space Telescope Science Institute at c h a p t e r 424 CHAPTER 4 The Law of Gravity Before 687, a large aount of data had been collected on the otions of the Moon and the planets, but a clear understanding of the forces causing these otions was not available. In that year, Isaac Newton provided the key that unlocked the secrets of the heavens. He knew, fro his first law, that a net force had to be acting on the Moon because without such a force the Moon would ove in a straight-line path rather than in its alost circular orbit. Newton reasoned that this force was the gravitational attraction eerted by the Earth on the Moon. He realized that the forces involved in the Earth Moon attraction and in the Sun planet attraction were not soething special to those systes, but rather were particular cases of a general and universal attraction between objects. In other words, Newton saw that the sae force of attraction that causes the Moon to follow its path around the Earth also causes an apple to fall fro a tree. As he put it, I deduced that the forces which keep the planets in their orbs ust be reciprocally as the squares of their distances fro the centers about which they revolve; and thereby copared the force requisite to keep the Moon in her orb with the force of gravity at the surface of the Earth; and found the answer pretty nearly. In this chapter we study the law of gravity. We place ephasis on describing the otion of the planets because astronoical data provide an iportant test of the validity of the law of gravity. We show that the laws of planetary otion developed by Johannes Kepler follow fro the law of gravity and the concept of conservation of angular oentu. We then derive a general epression for gravitational potential energy and eaine the energetics of planetary and satellite otion. We close by showing how the law of gravity is also used to deterine the force between a particle and an etended object. 4. NEWTN S LAW F UNIVERSAL GRAVITATIN You ay have heard the legend that Newton was struck on the head by a falling apple while napping under a tree. This alleged accident supposedly propted hi to iagine that perhaps all bodies in the Universe were attracted to each other in the sae way the apple was attracted to the Earth. Newton analyzed astronoical data on the otion of the Moon around the Earth. Fro that analysis, he ade the bold assertion that the force law governing the otion of planets was the sae as the force law that attracted a falling apple to the Earth. This was the first tie that earthly and heavenly otions were unified. We shall look at the atheatical details of Newton s analysis in Section 4.5. In 687 Newton published his work on the law of gravity in his treatise Matheatical Principles of Natural Philosophy. Newton s law of universal gravitation states that 4. Newton s Law of Universal Gravitation 4.2 Measuring the Gravitational Constant 4.3 Free-Fall Acceleration and the Gravitational Force 4.4 Kepler s Laws 4.5 The Law of Gravity and the Motion of Planets 4.6 The Gravitational Field 4.7 Gravitational Potential Energy 4.8 Energy Considerations in Planetary and Satellite Motion 4.9 (ptional) The Gravitational Force Between an Etended bject and a Particle 4.0 (ptional) The Gravitational Force Between a Particle and a Spherical Mass The law of gravity every particle in the Universe attracts every other particle with a force that is directly proportional to the product of their asses and inversely proportional to the square of the distance between the. If the particles have asses and 2 and are separated by a distance r, the agnitude of this gravitational force is F g G 2 r 2 (4.) 423

91 4. Newton s Law of Universal Gravitation CHAPTER 4 The Law of Gravity where G is a constant, called the universal gravitational constant, that has been easured eperientally. As noted in Eaple 6.6, its value in SI units is G N 2 /kg 2 (4.2) The for of the force law given by Equation 4. is often referred to as an inverse-square law because the agnitude of the force varies as the inverse square of the separation of the particles. We shall see other eaples of this type of force law in subsequent chapters. We can epress this force in vector for by defining a unit vector rˆ2 (Fig. 4.). Because this unit vector is directed fro particle to particle 2, the force eerted by particle on particle 2 is F 2 G 2 (4.3) r 2 rˆ2 where the inus sign indicates that particle 2 is attracted to particle, and hence the force ust be directed toward particle. By Newton s third law, the force eerted by particle 2 on particle, designated F 2, is equal in agnitude to F 2 and in the opposite direction. That is, these forces for an action reaction pair, and F 2 F 2. Several features of Equation 4.3 deserve ention. The gravitational force is a field force that always eists between two particles, regardless of the ediu that separates the. Because the force varies as the inverse square of the distance between the particles, it decreases rapidly with increasing separation. We can relate this fact to the geoetry of the situation by noting that the intensity of light eanating fro a point source drops off in the sae /r 2 anner, as shown in Figure 4.2. Another iportant point about Equation 4.3 is that the gravitational force eerted by a finite-size, spherically syetric ass distribution on a particle outside the distribution is the sae as if the entire ass of the distribution were concentrated at the center. For eaple, the force eerted by the r 2r Figure 4.2 Light radiating fro a point source drops off as /r 2, a relationship that atches the way the gravitational force depends on distance. When the distance fro the light source is doubled, the light has to cover four ties the area and thus is one fourth as bright. An inverse relationship between two quantities and y is one in which y k/, where k is a constant. A direct proportion between and y eists when y k. F 2 rˆ 2 Properties of the gravitational force QuickLab r F 2 2 Figure 4. The gravitational force between two particles is attractive. The unit vector rˆ2 is directed fro particle to particle 2. Note that F 2 F 2. Inflate a balloon just enough to for a sall sphere. Measure its diaeter. Use a arker to color in a -c square on its surface. Now continue inflating the balloon until it reaches twice the original diaeter. Measure the size of the square you have drawn. Also note how the color of the arked area has changed. Have you verified what is shown in Figure 4.2? EXAMPLE 4. Earth on a particle of ass near the Earth s surface has the agnitude F g G M E (4.4) R 2 E where M E is the Earth s ass and R E its radius. This force is directed toward the center of the Earth. We have evidence of the fact that the gravitational force acting on an object is directly proportional to its ass fro our observations of falling objects, discussed in Chapter 2. All objects, regardless of ass, fall in the absence of air resistance at the sae acceleration g near the surface of the Earth. According to Newton s second law, this acceleration is given by g F g /, where is the ass of the falling object. If this ratio is to be the sae for all falling objects, then F g ust be directly proportional to, so that the ass cancels in the ratio. If we consider the ore general situation of a gravitational force between any two objects with ass, such as two planets, this sae arguent can be applied to show that the gravitational force is proportional to one of the asses. We can choose either of the asses in the arguent, however; thus, the gravitational force ust be directly proportional to both asses, as can be seen in Equation MEASURING THE GRAVITATINAL CNSTANT The universal gravitational constant G was easured in an iportant eperient by Henry Cavendish (73 80) in 798. The Cavendish apparatus consists of two sall spheres, each of ass, fied to the ends of a light horizontal rod suspended by a fine fiber or thin etal wire, as illustrated in Figure 4.3. When two large spheres, each of ass M, are placed near the saller ones, the attractive force between saller and larger spheres causes the rod to rotate and twist the wire suspension to a new equilibriu orientation. The angle of rotation is easured by the deflection of a light bea reflected fro a irror attached to the vertical suspension. The deflection of the light is an effective technique for aplifying the otion. The eperient is carefully repeated with different asses at various separations. In addition to providing a value for G, the results show eperientally that the force is attractive, proportional to the product M, and inversely proportional to the square of the distance r. Mirror Billiards, Anyone? Three kg billiard balls are placed on a table at the corners of a right triangle, as shown in Figure 4.4. Calculate the gravitational force on the cue ball (designated ) resulting fro the other two balls. M r Light source Figure 4.3 Scheatic diagra of the Cavendish apparatus for easuring G. As the sall spheres of ass are attracted to the large spheres of ass M, the rod between the two sall spheres rotates through a sall angle. A light bea reflected fro a irror on the rotating apparatus easures the angle of rotation. The dashed line represents the original position of the rod. First we calculate separately the individual forces on the cue ball due to the other two balls, and then we find the vector su to get the resultant force. We can see graphically that this force should point upward and toward the

92 4.3 Free-Fall Acceleration and the Gravitational Force CHAPTER 4 The Law of Gravity right. We locate our coordinate aes as shown in Figure 4.4, placing our origin at the position of the cue ball N2 (0.300 kg)(0.300 kg) kg 2 (0.400 ) 2 j The force eerted by 2 on the cue ball is directed upward and is given by j N F 2 G This result shows that the gravitational forces between everyday objects have etreely sall agnitudes. The force e- 2 r 2 j 2 erted by 3 on the cue ball is directed to the right: 4.3 Figure y F 2 2 F F The resultant gravitational force acting on the cue ball is the vector su F 2 F 3. FREE-FALL ACCELERATIN AND THE GRAVITATINAL FRCE In Chapter 5, when defining g as the weight of an object of ass, we referred to g as the agnitude of the free-fall acceleration. Now we are in a position to obtain a ore fundaental description of g. Because the force acting on a freely falling object of ass near the Earth s surface is given by Equation 4.4, we can equate g to this force to obtain g G M E (4.5) R 2 E Now consider an object of ass located a distance h above the Earth s surface or a distance r fro the Earth s center, where r R E h. The agnitude of the gravitational force acting on this object is F g G M E r 2 g G M E R 2 E G M E (R E h) 2 Therefore, the resultant force on the cue ball is and the agnitude of this force is Eercise Find the direction of F. Answer i N F F 2 F 3 The gravitational force acting on the object at this position is also F g g, where g is the value of the free-fall acceleration at the altitude h. Substituting this epres- F 3 G 3 r 2 i N2 (0.300 kg)(0.300 kg) kg 2 (0.300 ) 2 i (3.75j 6.67i) 0 N F F 2 2 F 2 3 (3.75) 2 (6.67) N 29.3 counterclockwise fro the positive ais. Free-fall acceleration near the Earth s surface Variation of g with altitude EXAMPLE 4.2 sion for F g into the last equation shows that g is g GM E r 2 GM E (R E h) 2 (4.6) Thus, it follows that g decreases with increasing altitude. Because the weight of a body is g, we see that as r :, its weight approaches zero. Variation of g with Altitude h The International Space Station is designed to operate at an altitude of 350 k. When copleted, it will have a weight (easured at the Earth s surface) of N. What is its weight when in orbit? Because the station is above the surface of the Earth, we epect its weight in orbit to be less than its weight on Earth, N. Using Equation 4.6 with h 350 k, we obtain g GM E (R E h) 2 ( N 2 /kg 2 )( kg) ( ) /s 2 Because g/g 8.83/ , we conclude that the weight of the station at an altitude of 350 k is 90.% of the value at the Earth s surface. So the station s weight in orbit is (0.90)( N) N Values of g at other altitudes are listed in Table 4.. EXAMPLE 4.3 The Density of the Earth Using the fact that g 9.80 /s 2 at the Earth s surface, find the average density of the Earth. Using g 9.80 /s 2 and R E , we find fro Equation 4.5 that M E kg. Fro this result, and using the definition of density fro Chapter, we obtain V 4 3 3R kg 4 E 3( ) kg/ 3 TABLE 4. web The official web site for the International Space Station is Free-Fall Acceleration g at Various Altitudes Above the Earth s Surface Altitude h (k) g (/s 2 ) Because this value is about twice the density of ost rocks at the Earth s surface, we conclude that the inner core of the Earth has a density uch higher than the average value. It is ost aazing that the Cavendish eperient, which deterines G (and can be done on a tabletop), cobined with siple free-fall easureents of g, provides inforation about the core of the Earth.

93 4.4 Kepler s Laws CHAPTER 4 The Law of Gravity Kepler s analysis first showed that the concept of circular orbits around the Sun had to be abandoned. He eventually discovered that the orbit of Mars could be accurately described by an ellipse. Figure 4.5 shows the geoetric description of an ellipse. The longest diension is called the ajor ais and is of length 2a, where a is the seiajor ais. The shortest diension is the inor ais, of length 2b, where b is the seiinor ais. n either side of the center is a focal point, a distance c fro the center, where a 2 b 2 c 2. The Sun is located at one of the focal points of Mars s orbit. Kepler generalized his analysis to include the otions of all planets. The coplete analysis is suarized in three stateents known as Kepler s laws: Kepler s laws. All planets ove in elliptical orbits with the Sun at one focal point. 2. The radius vector drawn fro the Sun to a planet sweeps out equal areas in equal tie intervals. 3. The square of the orbital period of any planet is proportional to the cube of the seiajor ais of the elliptical orbit. Astronauts F. Story Musgrave and Jeffrey A. Hoffan, along with the Hubble Space Telescope and the space shuttle Endeavor, are all falling around the Earth. 4.4 KEPLER S LAWS People have observed the oveents of the planets, stars, and other celestial bodies for thousands of years. In early history, scientists regarded the Earth as the center of the Universe. This so-called geocentric odel was elaborated and foralized by the Greek astronoer Claudius Ptoley (c. 00 c. 70) in the second century A.D. and was accepted for the net 400 years. In 543 the Polish astronoer Nicolaus Copernicus ( ) suggested that the Earth and the other planets revolved in circular orbits around the Sun (the heliocentric odel). The Danish astronoer Tycho Brahe (546 60) wanted to deterine how the heavens were constructed, and thus he developed a progra to deterine the positions of both stars and planets. It is interesting to note that those observations of the planets and 777 stars visible to the naked eye were carried out with only a large setant and a copass. (The telescope had not yet been invented.) The Geran astronoer Johannes Kepler was Brahe s assistant for a short while before Brahe s death, whereupon he acquired his entor s astronoical data and spent 6 years trying to deduce a atheatical odel for the otion of the planets. Such data are difficult to sort out because the Earth is also in otion around the Sun. After any laborious calculations, Kepler found that Brahe s data on the revolution of Mars around the Sun provided the answer. Johannes Kepler Geran astronoer (57 630) The Geran astronoer Johannes Kepler is best known for developing the laws of planetary otion based on the careful observations of Tycho Brahe. (Art Resource) For ore inforation about Johannes Kepler, visit our Web site at F c a F 2 Figure 4.5 Plot of an ellipse. The seiajor ais has a length a, and the seiinor ais has a length b. The focal points are located at a distance c fro the center, where a 2 b 2 c 2. b Most of the planetary orbits are close to circular in shape; for eaple, the seiajor and seiinor aes of the orbit of Mars differ by only 0.4%. Mercury and Pluto have the ost elliptical orbits of the nine planets. In addition to the planets, there are any asteroids and coets orbiting the Sun that obey Kepler s laws. Coet Halley is such an object; it becoes visible when it is close to the Sun every 76 years. Its orbit is very elliptical, with a seiinor ais 76% saller than its seiajor ais. Although we do not prove it here, Kepler s first law is a direct consequence of the fact that the gravitational force varies as /r 2. That is, under an inverse-square gravitational-force law, the orbit of a planet can be shown atheatically to be an ellipse with the Sun at one focal point. Indeed, half a century after Kepler developed his laws, Newton deonstrated that these laws are a consequence of the gravitational force that eists between any two asses. Newton s law of universal gravitation, together with his developent of the laws of otion, provides the basis for a full atheatical solution to the otion of planets and satellites. 4.5 THE LAW F GRAVITY AND THE MTIN F PLANETS In forulating his law of gravity, Newton used the following reasoning, which supports the assuption that the gravitational force is proportional to the inverse square of the separation between the two interacting bodies. He copared the acceleration of the Moon in its orbit with the acceleration of an object falling near the Earth s surface, such as the legendary apple (Fig. 4.6). Assuing that both accelerations had the sae cause naely, the gravitational attraction of the Earth Newton used the inverse-square law to reason that the acceleration of the Moon toward the Earth (centripetal acceleration) should be proportional to /r 2 M, where r M is the distance between the centers of the Earth and the Moon. Furtherore, the acceleration of the apple toward the Earth should be proportional to /R 2 E, where R E is the radius of the Earth, or the distance between the centers of the Earth and the apple. Using the values r M and

94 4.5 The Law of Gravity and the Motion of Planets CHAPTER 4 The Law of Gravity v Kepler s Third Law R E , Newton predicted that the ratio of the Moon s acceleration a M to the apple s acceleration g would be a M g r M Moon (/r M) 2 (/R E ) 2 R E r M Therefore, the centripetal acceleration of the Moon is a M a M ( )(9.80 /s 2 ) /s 2 Newton also calculated the centripetal acceleration of the Moon fro a knowledge of its ean distance fro the Earth and its orbital period, T days s. In a tie T, the Moon travels a distance 2r M, which equals the circuference of its orbit. Therefore, its orbital speed is 2r M /T and its centripetal acceleration is a M v 2 (2r M /T) r M r M r M T ( ) ( s) 2 v g Figure 4.6 As it revolves around the Earth, the Moon eperiences a centripetal acceleration a M directed toward Earth the Earth. An object near the Earth s surface, such as the apple shown here, eperiences an acceleration g. (Diensions are not to R E scale.) Acceleration of the Moon Kepler s third law M S r M p Figure 4.7 A planet of ass M p oving in a circular orbit around the Sun. The orbits of all planets ecept Mercury and Pluto are nearly circular. It is inforative to show that Kepler s third law can be predicted fro the inversesquare law for circular orbits. 2 Consider a planet of ass M p oving around the Sun of ass M S in a circular orbit, as shown in Figure 4.7. Because the gravitational force eerted by the Sun on the planet is a radially directed force that keeps the planet oving in a circle, we can apply Newton s second law (F a) to the planet: GM S M p r 2 M pv 2 r Because the orbital speed v of the planet is siply 2r/T, where T is its period of revolution, the preceding epression becoes where K S is a constant given by GM S r 2 (2r/T) 2 r T GM S r 3 K S r 3 K S 4 2 GM S s 2 / 3 (4.7) Equation 4.7 is Kepler s third law. It can be shown that the law is also valid for elliptical orbits if we replace r with the length of the seiajor ais a. Note that the constant of proportionality K S is independent of the ass of the planet. Therefore, Equation 4.7 is valid for any planet. 3 Table 4.2 contains a collection of useful planetary data. The last colun verifies that T 2 /r 3 is a constant. The sall variations in the values in this colun reflect uncertainties in the easured values of the periods and seiajor aes of the planets. If we were to consider the orbit around the Earth of a satellite such as the Moon, then the proportionality constant would have a different value, with the Sun s ass replaced by the Earth s ass /s /s In other words, because the Moon is roughly 60 Earth radii away, the gravitational acceleration at that distance should be about /60 2 of its value at the Earth s surface. This is just the acceleration needed to account for the circular otion of the Moon around the Earth. The nearly perfect agreeent between this value and the value Newton obtained using g provides strong evidence of the inverse-square nature of the gravitational force law. Although these results ust have been very encouraging to Newton, he was deeply troubled by an assuption he ade in the analysis. To evaluate the acceleration of an object at the Earth s surface, Newton treated the Earth as if its ass were all concentrated at its center. That is, he assued that the Earth acted as a particle as far as its influence on an eterior object was concerned. Several years later, in 687, on the basis of his pioneering work in the developent of calculus, Newton proved that this assuption was valid and was a natural consequence of the law of universal gravitation. EXAMPLE 4.4 The Mass of the Sun Calculate the ass of the Sun using the fact that the period of the Earth s orbit around the Sun is s and its distance fro the Sun is Using Equation 4.7, we find that M S 4 2 r 3 GT 2 ( N 2 /kg 2 )( s) ( ) kg In Eaple 4.3, an understanding of gravitational forces enabled us to find out soething about the density of the Earth s core, and now we have used this understanding to deterine the ass of the Sun. 2 The orbits of all planets ecept Mercury and Pluto are very close to being circular; hence, we do not introduce uch error with this assuption. For eaple, the ratio of the seiinor ais to the seiajor ais for the Earth s orbit is b/a Equation 4.7 is indeed a proportion because the ratio of the two quantities T 2 and r 3 is a constant. The variables in a proportion are not required to be liited to the first power only.

95 4.5 The Law of Gravity and the Motion of Planets CHAPTER 4 The Law of Gravity TABLE 4.2 Useful Planetary Data Mean Period of Radius Revolution Mean Distance Body Mass (kg) () (s) fro Sun () Mercury Venus Earth Mars Jupiter Saturn Uranus Neptune Pluto Moon Sun Kepler s Second Law and Conservation of Angular Moentu Consider a planet of ass M p oving around the Sun in an elliptical orbit (Fig. 4.8). The gravitational force acting on the planet is always along the radius vector, directed toward the Sun, as shown in Figure 4.9a. When a force is directed toward or away fro a fied point and is a function of r only, it is called a central force. The torque acting on the planet due to this force is clearly zero; that is, because F is parallel to r, r F r F rˆ 0 (You ay want to revisit Section.2 to refresh your eory on the vector product.) Recall fro Equation.9, however, that torque equals the tie rate of change of angular oentu: Therefore, because the gravitational d L/dt. Separate views of Jupiter and of Periodic Coet Shoeaker Levy 9 both taken with the Hubble Space Telescope about two onths before Jupiter and the coet collided in July 994 were put together with the use of a coputer. Their relative sizes and distances were altered. The black spot on Jupiter is the shadow of its oon Io. T 2 r 3 (s2 / 3 ) D C Sun S A B Figure 4.8 Kepler s second law is called the law of equal areas. When the tie interval required for a planet to travel fro A to B is equal to the tie interval required for it to go fro C to D, the two areas swept out by the planet s radius vector are equal. Note that in order for this to be true, the planet ust be oving faster between C and D than between A and B. Sun M S Sun r F g (a) r (b) da M p EXAMPLE 4.5 v dr = vdt Figure 4.9 (a) The gravitational force acting on a planet is directed toward the Sun, along the radius vector. (b) As a planet orbits the Sun, the area swept out by the radius vector in a tie dt is equal to one-half the area of the parallelogra fored by the vectors r and d r vdt. force eerted by the Sun on a planet results in no torque on the planet, the angular oentu L of the planet is constant: L r p r M p v M p r v constant (4.8) Because L reains constant, the planet s otion at any instant is restricted to the plane fored by r and v. We can relate this result to the following geoetric consideration. The radius vector r in Figure 4.9b sweeps out an area da in a tie dt. This area equals onehalf the area r dr of the parallelogra fored by the vectors r and dr (see Section.2). Because the displaceent of the planet in a tie dt is dr vdt, we can say that da 2 r dr 2 r v dt L dt 2M p da dt L constant 2M p where L and M p are both constants. Thus, we conclude that (4.9) the radius vector fro the Sun to a planet sweeps out equal areas in equal tie intervals. It is iportant to recognize that this result, which is Kepler s second law, is a consequence of the fact that the force of gravity is a central force, which in turn iplies that angular oentu is constant. Therefore, Kepler s second law applies to any situation involving a central force, whether inverse-square or not. Motion in an Elliptical rbit A satellite of ass oves in an elliptical orbit around the Earth (Fig. 4.0). The iniu distance of the satellite fro the Earth is called the perigee (indicated by p in Fig. Figure 4.0 v a r p a p r a As a satellite oves around the Earth in an elliptical orbit, its angular oentu is constant. Therefore, v a r a v p r p, where the subscripts a and p represent apogee and perigee, respectively. v p 4.0), and the aiu distance is called the apogee (indicated by a). If the speed of the satellite at p is v p, what is its speed at a? As the satellite oves fro perigee toward apogee, it is oving farther fro the Earth. Thus, a coponent of the gravitational force eerted by the Earth on the satellite is opposite the velocity vector. Negative work is done on the satellite, which causes it to slow down, according to the work kinetic energy theore. As a result, we epect the speed at apogee to be lower than the speed at perigee. The angular oentu of the satellite relative to the Earth is r v r v. At the points a and p, v is perpendicular to r. Therefore, the agnitude of the angular oentu at these positions is L a v a r a and L p v p r p. Because angular oentu is constant, we see that v a r a v p r p r p v a v r p a

96 4.6 The Gravitational Field CHAPTER 4 The Law of Gravity Quick Quiz 4. How would you eplain the fact that Saturn and Jupiter have periods uch greater than one year? 4.6 THE GRAVITATINAL FIELD When Newton published his theory of universal gravitation, it was considered a success because it satisfactorily eplained the otion of the planets. Since 687 the sae theory has been used to account for the otions of coets, the deflection of a Cavendish balance, the orbits of binary stars, and the rotation of galaies. Nevertheless, both Newton s conteporaries and his successors found it difficult to accept the concept of a force that acts through a distance, as entioned in Section 5.. They asked how it was possible for two objects to interact when they were not in contact with each other. Newton hiself could not answer that question. An approach to describing interactions between objects that are not in contact cae well after Newton s death, and it enables us to look at the gravitational interaction in a different way. As described in Section 5., this alternative approach uses the concept of a gravitational field that eists at every point in space. When a particle of ass is placed at a point where the gravitational field is g, the particle eperiences a force F g g. In other words, the field eerts a force on the particle. Hence, the gravitational field g is defined as g F g (4.0) That is, the gravitational field at a point in space equals the gravitational force eperienced by a test particle placed at that point divided by the ass of the test particle. Notice that the presence of the test particle is not necessary for the field to eist the Earth creates the gravitational field. We call the object creating the field the source particle (although the Earth is clearly not a particle; we shall discuss shortly the fact that we can approiate the Earth as a particle for the purpose of finding the gravitational field that it creates). We can detect the presence of the field and easure its strength by placing a test particle in the field and noting the force eerted on it. Although the gravitational force is inherently an interaction between two objects, the concept of a gravitational field allows us to factor out the ass of one of the objects. In essence, we are describing the effect that any object (in this case, the Earth) has on the epty space around itself in ters of the force that would be present if a second object were soewhere in that space. 4 As an eaple of how the field concept works, consider an object of ass near the Earth s surface. Because the gravitational force acting on the object has a agnitude GM E /r 2 (see Eq. 4.4), the field g at a distance r fro the center of the Earth is g F g (4.) GM E r 2 rˆ where rˆ is a unit vector pointing radially outward fro the Earth and the inus 4 We shall return to this idea of ass affecting the space around it when we discuss Einstein s theory of gravitation in Chapter 39. Gravitational field sign indicates that the field points toward the center of the Earth, as illustrated in Figure 4.a. Note that the field vectors at different points surrounding the Earth vary in both direction and agnitude. In a sall region near the Earth s surface, the downward field g is approiately constant and unifor, as indicated in Figure 4.b. Equation 4. is valid at all points outside the Earth s surface, assuing that the Earth is spherical. At the Earth s surface, where r R E, g has a agnitude of 9.80 N/kg. 4.7 (a) Figure 4. (a) The gravitational field vectors in the vicinity of a unifor spherical ass such as the Earth vary in both direction and agnitude. The vectors point in the direction of the acceleration a particle would eperience if it were placed in the field. The agnitude of the field vector at any location is the agnitude of the free-fall acceleration at that location. (b) The gravitational field vectors in a sall region near the Earth s surface are unifor in both direction and agnitude. GRAVITATINAL PTENTIAL ENERGY In Chapter 8 we introduced the concept of gravitational potential energy, which is the energy associated with the position of a particle. We ephasized that the gravitational potential energy function U gy is valid only when the particle is near the Earth s surface, where the gravitational force is constant. Because the gravitational force between two particles varies as /r 2, we epect that a ore general potential energy function one that is valid without the restriction of having to be near the Earth s surface will be significantly different fro U gy. Before we calculate this general for for the gravitational potential energy function, let us first verify that the gravitational force is conservative. (Recall fro Section 8.2 that a force is conservative if the work it does on an object oving between any two points is independent of the path taken by the object.) To do this, we first note that the gravitational force is a central force. By definition, a central force is any force that is directed along a radial line to a fied center and has a agnitude that depends only on the radial coordinate r. Hence, a central force can be represented by F(r)rˆ, where rˆ is a unit vector directed fro the origin to the particle, as shown in Figure 4.2. Consider a central force acting on a particle oving along the general path P to Q in Figure 4.2. The path fro P to Q can be approiated by a series of (b)

97 4.7 Gravitational Potential Energy CHAPTER 4 The Law of Gravity steps according to the following procedure. In Figure 4.2, we draw several thin wedges, which are shown as dashed lines. The outer boundary of our set of wedges is a path consisting of short radial line segents and arcs (gray in the figure). We select the length of the radial diension of each wedge such that the short arc at the wedge s wide end intersects the actual path of the particle. Then we can approiate the actual path with a series of zigzag oveents that alternate between oving along an arc and oving along a radial line. By definition, a central force is always directed along one of the radial segents; therefore, the work done by F along any radial segent is dw F dr F(r) dr You should recall that, by definition, the work done by a force that is perpendicular to the displaceent is zero. Hence, the work done in oving along any arc is zero because F is perpendicular to the displaceent along these segents. Therefore, the total work done by F is the su of the contributions along the radial segents: W rf ri F(r) dr where the subscripts i and f refer to the initial and final positions. Because the integrand is a function only of the radial position, this integral depends only on the initial and final values of r. Thus, the work done is the sae over any path fro P to Q. Because the work done is independent of the path and depends only on the end points, we conclude that any central force is conservative. We are now assured that a potential energy function can be obtained once the for of the central force is specified. Recall fro Equation 8.2 that the change in the gravitational potential energy associated with a given displaceent is defined as the negative of the work done by the gravitational force during that displaceent: U U f U i rf F(r) dr (4.2) We can use this result to evaluate the gravitational potential energy function. Consider a particle of ass oving between two points P and Q above the Earth s surface (Fig. 4.3). The particle is subject to the gravitational force given by Equation 4.. We can epress this force as F(r) GM E r 2 where the negative sign indicates that the force is attractive. Substituting this epression for F(r) into Equation 4.2, we can copute the change in the gravita- R E M E r i P F g r f F g Q ri Figure 4.3 As a particle of ass oves fro P to Q above the Earth s surface, the gravitational potential energy changes according to Equation 4.2. Work done by a central force rˆ r i P rˆ F r f Radial segent Q Arc Figure 4.2 A particle oves fro P to Q while acted on by a central force F, which is directed radially. The path is broken into a series of radial segents and arcs. Because the work done along the arcs is zero, the work done is independent of the path and depends only on r f and r i. Change in gravitational potential energy Gravitational potential energy of the Earth particle syste for r R E M E U GM E R E Earth R E Figure 4.4 Graph of the gravitational potential energy U versus r for a particle above the Earth s surface. The potential energy goes to zero as r approaches infinity. r 2 r 3 2 r 23 Figure 4.5 Three interacting particles. 3 r tional potential energy function: U f U i GM E rf ri U f U i GM E r f r i (4.3) As always, the choice of a reference point for the potential energy is copletely arbitrary. It is custoary to choose the reference point where the force is zero. Taking U i 0 at r i, we obtain the iportant result U GM E (4.4) r This epression applies to the Earth particle syste where the two asses are separated by a distance r, provided that r R E. The result is not valid for particles inside the Earth, where r R E. (The situation in which r R E is treated in Section 4.0.) Because of our choice of U i, the function U is always negative (Fig. 4.4). Although Equation 4.4 was derived for the particle Earth syste, it can be applied to any two particles. That is, the gravitational potential energy associated with any pair of particles of asses and 2 separated by a distance r is U G 2 r dr r 2 GM E r rf r i (4.5) This epression shows that the gravitational potential energy for any pair of particles varies as /r, whereas the force between the varies as /r 2. Furtherore, the potential energy is negative because the force is attractive and we have taken the potential energy as zero when the particle separation is infinite. Because the force between the particles is attractive, we know that an eternal agent ust do positive work to increase the separation between the. The work done by the eternal agent produces an increase in the potential energy as the two particles are separated. That is, U becoes less negative as r increases. When two particles are at rest and separated by a distance r, an eternal agent has to supply an energy at least equal to G 2 /r in order to separate the particles to an infinite distance. It is therefore convenient to think of the absolute value of the potential energy as the binding energy of the syste. If the eternal agent supplies an energy greater than the binding energy, the ecess energy of the syste will be in the for of kinetic energy when the particles are at an infinite separation. We can etend this concept to three or ore particles. In this case, the total potential energy of the syste is the su over all pairs of particles. 5 Each pair contributes a ter of the for given by Equation 4.5. For eaple, if the syste contains three particles, as in Figure 4.5, we find that U total U 2 U 3 U 23 G (4.6) r 2 r 3 r 23 The absolute value of U total represents the work needed to separate the particles by an infinite distance. 5 The fact that potential energy ters can be added for all pairs of particles stes fro the eperiental fact that gravitational forces obey the superposition principle.

98 4.8 Energy Considerations in Planetary and Satellite Motion CHAPTER 4 The Law of Gravity EXAMPLE The Change in Potential Energy A particle of ass is displaced through a sall vertical distance y near the Earth s surface. Show that in this situation the general epression for the change in gravitational potential energy given by Equation 4.3 reduces to the failiar relationship U g y. We can epress Equation 4.3 in the for U GM E r f r i GM E r f r i r i r f ENERGY CNSIDERATINS IN PLANETARY AND SATELLITE MTIN Consider a body of ass oving with a speed v in the vicinity of a assive body of ass M, where M W. The syste ight be a planet oving around the Sun, a satellite in orbit around the Earth, or a coet aking a one-tie flyby of the Sun. If we assue that the body of ass M is at rest in an inertial reference frae, then the total echanical energy E of the two-body syste when the bodies are separated by a distance r is the su of the kinetic energy of the body of ass and the potential energy of the syste, given by Equation 4.5: 6 E K U E 2 v 2 GM r (4.7) This equation shows that E ay be positive, negative, or zero, depending on the value of v. However, for a bound syste, 7 such as the Earth Sun syste, E is necessarily less than zero because we have chosen the convention that U : 0 as r :. We can easily establish that E 0 for the syste consisting of a body of ass oving in a circular orbit about a body of ass M W (Fig. 4.6). Newton s second law applied to the body of ass gives GM r 2 a v 2 r 6 You ight recognize that we have ignored the acceleration and kinetic energy of the larger body. To see that this siplification is reasonable, consider an object of ass falling toward the Earth. Because the center of ass of the object Earth syste is effectively stationary, it follows that v M E v E. Thus, the Earth acquires a kinetic energy equal to 2 M Ev 2 E 2 2 v 2 M E where K is the kinetic energy of the object. Because M E W, this result shows that the kinetic energy of the Earth is negligible. 7 f the three eaples provided at the beginning of this section, the planet oving around the Sun and a satellite in orbit around the Earth are bound systes the Earth will always stay near the Sun, and the satellite will always stay near the Earth. The one-tie coet flyby represents an unbound syste the coet interacts once with the Sun but is not bound to it. Thus, in theory the coet can ove infinitely far away fro the Sun. M E K If both the initial and final positions of the particle are close to the Earth s surface, then r and r i r f R 2 f r i y E. (Recall that r is easured fro the center of the Earth.) Therefore, the change in potential energy becoes U GM E R 2 y g y E where we have used the fact that g GM E /R 2 E (Eq. 4.5). Keep in ind that the reference point is arbitrary because it is the change in potential energy that is eaningful. v M Figure 4.6 A body of ass oving in a circular orbit about a uch larger body of ass M. r Total energy for circular orbits EXAMPLE 4.7 Multiplying both sides by r and dividing by 2 gives 2 v 2 GM 2r Substituting this into Equation 4.7, we obtain E GM 2r E GM 2r GM r (4.8) (4.9) This result clearly shows that the total echanical energy is negative in the case of circular orbits. Note that the kinetic energy is positive and equal to one-half the absolute value of the potential energy. The absolute value of E is also equal to the binding energy of the syste, because this aount of energy ust be provided to the syste to ove the two asses infinitely far apart. The total echanical energy is also negative in the case of elliptical orbits. The epression for E for elliptical orbits is the sae as Equation 4.9 with r replaced by the seiajor ais length a. Furtherore, the total energy is constant if we assue that the syste is isolated. Therefore, as the body of ass oves fro P to Q in Figure 4.3, the total energy reains constant and Equation 4.7 gives E 2 v i 2 GM 2 (4.20) r v f 2 GM i r f Cobining this stateent of energy conservation with our earlier discussion of conservation of angular oentu, we see that both the total energy and the total angular oentu of a gravitationally bound, two-body syste are constants of the otion. Changing the rbit of a Satellite The space shuttle releases a 470-kg counications satellite while in an orbit that is 280 k above the surface of the Earth. A rocket engine on the satellite boosts it into a geosynchronous orbit, which is an orbit in which the satellite stays directly over a single location on the Earth. How uch energy did the engine have to provide? First we ust deterine the radius of a geosynchronous orbit. Then we can calculate the change in energy needed to boost the satellite into orbit. The period of the orbit T ust be one day ( s), so that the satellite travels once around the Earth in the sae tie that the Earth spins once on its ais. Knowing the period, we can then apply Kepler s third law (Eq. 4.7) to find the radius, once we replace K S with K E 4 2 /GM E s 2 / 3 : T 2 K E r 3 3 T r 2 3 ( s) K E s 2 / R f This is a little ore than i above the Earth s surface. We ust also deterine the initial radius (not the altitude above the Earth s surface) of the satellite s orbit when it was still in the shuttle s cargo bay. This is siply Now, applying Equation 4.9, we obtain, for the total initial and final energies, The energy required fro the engine to boost the satellite is E engine E f E i GM E 2 R f R i ( N 2 /kg 2 )( kg)(470 kg) 2 R E 280 k R i E i GM E 2R i J E f GM E 2R f

99 4.8 Energy Considerations in Planetary and Satellite Motion CHAPTER 4 The Law of Gravity This is the energy equivalent of 89 gal of gasoline. NASA engineers ust account for the changing ass of the spacecraft as it ejects burned fuel, soething we have not done here. Would you epect the calculation that includes the effect of this changing ass to yield a greater or lesser aount of energy required fro the engine? If we wish to deterine how the energy is distributed after the engine is fired, we find fro Equation 4.8 that the change in kinetic energy is K (GM E /2)(/R f /R i ) J (a decrease), and the corresponding change in potential energy is U GM E (/R f /R i ) J (an increase). Thus, the change in echanical energy of the syste is E K U J, as we already calculated. The firing of the engine results in an increase in the total echanical energy of the syste. Because an increase in potential energy is accopanied by a decrease in kinetic energy, we conclude that the speed of an orbiting satellite decreases as its altitude increases. EXAMPLE 4.8 Escape Speed of a Rocket Calculate the escape speed fro the Earth for a kg spacecraft, and deterine the kinetic energy it ust have at the Earth s surface in order to escape the Earth s gravitational field. v esc 2GM E R E Using Equation 4.22 gives This corresponds to about i/h. The kinetic energy of the spacecraft is /s K 2 v 2 esc 2 ( kg)( /s) J 2( N 2 /kg 2 )( kg) This is equivalent to about gal of gasoline. Escape Speed Suppose an object of ass is projected vertically upward fro the Earth s surface with an initial speed v i, as illustrated in Figure 4.7. We can use energy considerations to find the iniu value of the initial speed needed to allow the object to escape the Earth s gravitational field. Equation 4.7 gives the total energy of the object at any point. At the surface of the Earth, v v i and r r i R E. When the object reaches its aiu altitude, v v f 0 and r r f r a. Because the total energy of the syste is constant, substituting these conditions into Equation 4.20 gives Solving for v 2 i gives 2 v i 2 GM E GM E R E r a v 2 i 2GM E R E r a (4.2) Therefore, if the initial speed is known, this epression can be used to calculate the aiu altitude h because we know that h r a R E We are now in a position to calculate escape speed, which is the iniu speed the object ust have at the Earth s surface in order to escape fro the influence of the Earth s gravitational field. Traveling at this iniu speed, the object continues to ove farther and farther away fro the Earth as its speed asyptotically approaches zero. Letting r a : in Equation 4.2 and taking v i v esc, we obtain R E v f = 0 v i M E h r a Figure 4.7 An object of ass projected upward fro the Earth s surface with an initial speed v i reaches a aiu altitude h. TABLE 4.3 Escape Speeds fro the Surfaces of the Planets, Moon, and Sun Body v esc (k/s) Mercury 4.3 Venus 0.3 Earth.2 Moon 2.3 Mars 5.0 Jupiter 60 Saturn 36 Uranus 22 Neptune 24 Pluto. Sun 68 Equations 4.2 and 4.22 can be applied to objects projected fro any planet. That is, in general, the escape speed fro the surface of any planet of ass M and radius R is Escape speeds for the planets, the Moon, and the Sun are provided in Table 4.3. Note that the values vary fro. k/s for Pluto to about 68 k/s for the Sun. These results, together with soe ideas fro the kinetic theory of gases (see Chapter 2), eplain why soe planets have atospheres and others do not. As we shall see later, a gas olecule has an average kinetic energy that depends on the teperature of the gas. Hence, lighter olecules, such as hydrogen and heliu, have a higher average speed than heavier species at the sae teperature. When the average speed of the lighter olecules is not uch less than the escape speed of a planet, a significant fraction of the have a chance to escape fro the planet. This echanis also eplains why the Earth does not retain hydrogen olecules and heliu atos in its atosphere but does retain heavier olecules, such as oygen and nitrogen. n the other hand, the very large escape speed for Jupiter enables that planet to retain hydrogen, the priary constituent of its atosphere. Quick Quiz 4.2 v esc 2GM R v esc 2GM E R E (4.22) Escape speed If you were a space prospector and discovered gold on an asteroid, it probably would not be a good idea to jup up and down in eciteent over your find. Why? Note that this epression for v esc is independent of the ass of the object. In other words, a spacecraft has the sae escape speed as a olecule. Furtherore, the result is independent of the direction of the velocity and ignores air resistance. If the object is given an initial speed equal to v esc, its total energy is equal to zero. This can be seen by noting that when r :, the object s kinetic energy and its potential energy are both zero. If v i is greater than v esc, the total energy is greater than zero and the object has soe residual kinetic energy as r :. Quick Quiz 4.3 Figure 4.8 is a drawing by Newton showing the path of a stone thrown fro a ountaintop. He shows the stone landing farther and farther away when thrown at higher and higher speeds (at points D, E, F, and G), until finally it is thrown all the way around the Earth. Why didn t Newton show the stone landing at B and A before it was going fast enough to coplete an orbit?

100 4.9 The Gravitational Force Between an Etended bject and a Particle CHAPTER 4 The Law of Gravity ptional Section 4.9 Figure 4.8 The greater the velocity...with which [a stone] is projected, the farther it goes before it falls to the Earth. We ay therefore suppose the velocity to be so increased, that it would describe an arc of, 2, 5, 0, 00, 000 iles before it arrived at the Earth, till at last, eceeding the liits of the Earth, it should pass into space without touching. Sir Isaac Newton, Syste of the World. THE GRAVITATINAL FRCE BETWEEN AN EXTENDED BJECT AND A PARTICLE We have ephasized that the law of universal gravitation given by Equation 4.3 is valid only if the interacting objects are treated as particles. In view of this, how can we calculate the force between a particle and an object having finite diensions? This is accoplished by treating the etended object as a collection of particles and aking use of integral calculus. We first evaluate the potential energy function, and then calculate the gravitational force fro that function. We obtain the potential energy associated with a syste consisting of a particle of ass and an etended object of ass M by dividing the object into any eleents, each having a ass M i (Fig. 4.9). The potential energy associated with the syste consisting of any one eleent and the particle is U G M i /r i, where r i is the distance fro the particle to the eleent M i. The total potential energy of the overall syste is obtained by taking the su over all eleents as M i : 0. In this liit, we can epress U in integral for as U G dm (4.23) r nce U has been evaluated, we obtain the force eerted by the etended object on the particle by taking the negative derivative of this scalar function (see Section 8.6). If the etended object has spherical syetry, the function U depends only on r, and the force is given by du/dr. We treat this situation in Section 4.0. In principle, one can evaluate U for any geoetry; however, the integration can be cubersoe. An alternative approach to evaluating the gravitational force between a particle and an etended object is to perfor a vector su over all ass eleents of the object. Using the procedure outlined in evaluating U and the law of universal gravitation in the for shown in Equation 4.3, we obtain, for the total force eerted on the particle F g G dm (4.24) r 2 rˆ where rˆ is a unit vector directed fro the eleent dm toward the particle (see Fig. 4.9) and the inus sign indicates that the direction of the force is opposite that of rˆ. This procedure is not always recoended because working with a vector function is ore difficult than working with the scalar potential energy function. However, if the geoetry is siple, as in the following eaple, the evaluation of F can be straightforward. M M i Figure 4.9 A particle of ass interacting with an etended object of ass M. The total gravitational force eerted by the object on the particle can be obtained by dividing the object into nuerous eleents, each having a ass M i, and then taking a vector su over the forces eerted by all eleents. Total force eerted on a particle by an etended object rˆ r i EXAMPLE 4.9 Gravitational Force Between a Particle and a Bar The left end of a hoogeneous bar of length L and ass M is at a distance h fro a particle of ass (Fig. 4.20). Calculate the total gravitational force eerted by the bar on the particle. The arbitrary segent of the bar of length d has a ass dm. Because the ass per unit length is constant, it follows that the ratio of asses dm/m is equal to the ratio Figure 4.20 Force on a particle due to a spherical shell y h The gravitational force eerted by the bar on the particle is directed to the right. Note that the bar is not equivalent to a particle of ass M located at the center of ass of the bar. L d ptional Section 4.0 of lengths d/l, and so dm (M/L) d. In this proble, the variable r in Equation 4.24 is the distance shown in Figure 4.20, the unit vector rˆ is rˆ i, and the force acting on the particle is to the right; therefore, Equation 4.24 gives us F g G hl h F g GM L hl GM h(h L) i We see that the force eerted on the particle is in the positive direction, which is what we epect because the gravitational force is attractive. Note that in the liit L : 0, the force varies as /h 2, which is what we epect for the force between two point asses. Furtherore, if h W L, the force also varies as /h 2. This can be seen by noting that the denoinator of the epression for F g can be epressed in the for h 2 ( L/h), which is approiately equal to h 2 when h W L. Thus, when bodies are separated by distances that are great relative to their characteristic diensions, they behave like particles. THE GRAVITATINAL FRCE BETWEEN A PARTICLE AND A SPHERICAL MASS We have already stated that a large sphere attracts a particle outside it as if the total ass of the sphere were concentrated at its center. We now describe the force acting on a particle when the etended object is either a spherical shell or a solid sphere, and then apply these facts to soe interesting systes. Spherical Shell Case. If a particle of ass is located outside a spherical shell of ass M at, for instance, point P in Figure 4.2a, the shell attracts the particle as though the ass of the shell were concentrated at its center. We can show this, as Newton did, with integral calculus. Thus, as far as the gravitational force acting on a particle outside the shell is concerned, a spherical shell acts no differently fro the solid spherical distributions of ass we have seen. Case 2. If the particle is located inside the shell (at point P in Fig. 4.2b), the gravitational force acting on it can be shown to be zero. We can epress these two iportant results in the following way: F g GM (4.25a) r 2 rˆ for r R F g 0 for r R (4.25b) The gravitational force as a function of the distance r is plotted in Figure 4.2c. Md L h 2 (i) G M hl d L h 2 i i

101 4.0 The Gravitational Force Between a Particle and a Spherical Mass CHAPTER 4 The Law of Gravity M Q F QP P ass of the sphere were concentrated at its center. We have used this notion at several places in this chapter already, and we can argue it fro Equation 4.25a. A solid sphere can be considered to be a collection of concentric spherical shells. The asses of all of the shells can be interpreted as being concentrated at their coon center, and the gravitational force is equivalent to that due to a particle of ass M located at that center. M (a) F Top, P P Q F Q P Force on a particle due to a solid sphere Case 2. If a particle of ass is located inside a hoogeneous solid sphere of ass M (at point Q in Fig. 4.22), the gravitational force acting on it is due only to the ass M contained within the sphere of radius r R, shown in Figure In other words, F g GM r 2 rˆ for r R (4.26a) F g GM (4.26b) r 2 rˆ for r R This also follows fro spherical-shell Case because the part of the sphere that is F Botto, P (b) M R M F g P F g r Q R r (c) Figure 4.2 (a) The nonradial coponents of the gravitational forces eerted on a particle of ass located at point P outside a spherical shell of ass M cancel out. (b) The spherical shell can be broken into rings. Even though point P is closer to the top ring than to the botto ring, the botto ring is larger, and the gravitational forces eerted on the particle at P by the atter in the two rings cancel each other. Thus, for a particle located at any point P inside the shell, there is no gravitational force eerted on the particle by the ass M of the shell. (c) The agnitude of the gravitational force versus the radial distance r fro the center of the shell. F g The shell does not act as a gravitational shield, which eans that a particle inside a shell ay eperience forces eerted by bodies outside the shell. R r Solid Sphere Case. If a particle of ass is located outside a hoogeneous solid sphere of ass M (at point P in Fig. 4.22), the sphere attracts the particle as though the Figure 4.22 The gravitational force acting on a particle when it is outside a unifor solid sphere is GM/r 2 and is directed toward the center of the sphere. The gravitational force acting on the particle when it is inside such a sphere is proportional to r and goes to zero at the center.

102 4.0 The Gravitational Force Between a Particle and a Spherical Mass CHAPTER 4 The Law of Gravity farther fro the center than Q can be treated as a series of concentric spherical shells that do not eert a net force on the particle because the particle is inside the. Because the sphere is assued to have a unifor density, it follows that the ratio of asses M/M is equal to the ratio of volues V/V, where V is the total volue of the sphere and V is the volue within the sphere of radius r only: 4 M V M V 3r 3 4 3R 3 r 3 R 3 Solving this equation for M and substituting the value obtained into Equation 4.26b, we have F g GM (4.27) R 3 r rˆ for r R This equation tells us that at the center of the solid sphere, where r 0, the gravitational force goes to zero, as we intuitively epect. The force as a function of r is plotted in Figure Case 3. If a particle is located inside a solid sphere having a density that is spherically syetric but not unifor, then M in Equation 4.26b is given by an integral of the for M dv, where the integration is taken over the volue contained within the sphere of radius r in Figure We can evaluate this integral if the radial variation of is given. In this case, we take the volue eleent dv as the volue of a spherical shell of radius r and thickness dr, and thus dv 4r 2 dr. For eaple, if Ar, where A is a constant, it is left to a proble (Proble 63) to show that M Ar 4. Hence, we see fro Equation 4.26b that F is proportional to r 2 in this case and is zero at the center. Figure 4.23 y F g r θ An object oves along a tunnel dug through the Earth. The coponent of the gravitational force F g along the ais is the driving force for the otion. Note that this coponent always acts toward. Solving for a, we obtain If we use the sybol 2 for the coefficient of GM E /R 3 E 2 we see that () a 2 a GM E R 3 E an epression that atches the atheatical for of Equation 3.9, which gives the acceleration of a particle in siple haronic otion: a 2. Therefore, Equation (), which we have derived for the acceleration of our object in the tunnel, is the acceleration equation for siple haronic otion at angular speed with Thus, the object in the tunnel oves in the sae way as a block hanging fro a spring! The period of oscillation is T 2 R 3 2 E GM E s GM E 84.3 in R 3 E 2 ( ) 3 ( N 2 /kg 2 )( kg) This period is the sae as that of a satellite traveling in a circular orbit just above the Earth s surface (ignoring any trees, buildings, or other objects in the way). Note that the result is independent of the length of the tunnel. A proposal has been ade to operate a ass-transit syste between any two cities, using the principle described in this eaple. A one-way trip would take about 42 in. A ore precise calculation of the otion ust account for the fact that the Earth s density is not unifor. More iportant, there are any practical probles to consider. For instance, it would be ipossible to achieve a frictionless tunnel, and so soe auiliary power source would be required. Can you think of other probles? Quick Quiz 4.4 A particle is projected through a sall hole into the interior of a spherical shell. Describe the otion of the particle inside the shell. EXAMPLE 4.0 A Free Ride, Thanks to Gravity An object of ass oves in a sooth, straight tunnel dug between two points on the Earth s surface (Fig. 4.23). Show that the object oves with siple haronic otion, and find the period of its otion. Assue that the Earth s density is unifor. The gravitational force eerted on the object acts toward the Earth s center and is given by Equation 4.27: F g GM R 3 r rˆ We receive our first indication that this force should result in siple haronic otion by coparing it to Hooke s law, first seen in Section 7.3. Because the gravitational force on the object is linearly proportional to the displaceent, the object eperiences a Hooke s law force. The y coponent of the gravitational force on the object is balanced by the noral force eerted by the tunnel wall, and the coponent is F GM E R 3 r cos E Because the coordinate of the object is r cos, we can write F GM E R 3 E Applying Newton s second law to the otion along the direction gives F GM E R 3 a E SUMMARY Newton s law of universal gravitation states that the gravitational force of attraction between any two particles of asses and 2 separated by a distance r has the agnitude F g G 2 (4.) r 2 where G N 2 /kg 2 is the universal gravitational constant. This equation enables us to calculate the force of attraction between asses under a wide variety of circustances. An object at a distance h above the Earth s surface eperiences a gravitational force of agnitude g, where g is the free-fall acceleration at that elevation: g GM E r 2 GM E (R E h) 2 (4.6)

103 Suary CHAPTER 4 The Law of Gravity In this epression, M E is the ass of the Earth and R E is its radius. Thus, the weight of an object decreases as the object oves away fro the Earth s surface. Kepler s laws of planetary otion state that. All planets ove in elliptical orbits with the Sun at one focal point. 2. The radius vector drawn fro the Sun to a planet sweeps out equal areas in equal tie intervals. 3. The square of the orbital period of any planet is proportional to the cube of the seiajor ais of the elliptical orbit. Kepler s third law can be epressed as T GM S r 3 (4.7) where M S is the ass of the Sun and r is the orbital radius. For elliptical orbits, Equation 4.7 is valid if r is replaced by the seiajor ais a. Most planets have nearly circular orbits around the Sun. The gravitational field at a point in space equals the gravitational force eperienced by any test particle located at that point divided by the ass of the test particle: g F g (4.0) The gravitational force is conservative, and therefore a potential energy function can be defined. The gravitational potential energy associated with two particles separated by a distance r is U G 2 (4.5) r where U is taken to be zero as r :. The total potential energy for a syste of particles is the su of energies for all pairs of particles, with each pair represented by a ter of the for given by Equation 4.5. If an isolated syste consists of a particle of ass oving with a speed v in the vicinity of a assive body of ass M, the total energy E of the syste is the su of the kinetic and potential energies: E 2 v 2 GM (4.7) r The total energy is a constant of the otion. If the particle oves in a circular orbit of radius r around the assive body and if M W, the total energy of the syste is E GM (4.9) 2r The total energy is negative for any bound syste. The escape speed for an object projected fro the surface of the Earth is v esc 2GM E R E (4.22) QUESTINS. Use Kepler s second law to convince yourself that the Earth ust ove faster in its orbit during Deceber, when it is closest to the Sun, than during June, when it is farthest fro the Sun. 2. The gravitational force that the Sun eerts on the Moon is about twice as great as the gravitational force that the Earth eerts on the Moon. Why doesn t the Sun pull the Moon away fro the Earth during a total eclipse of the Sun? 3. If a syste consists of five particles, how any ters appear in the epression for the total potential energy? How any ters appear if the syste consists of N particles? 4. Is it possible to calculate the potential energy function associated with a particle and an etended body without knowing the geoetry or ass distribution of the etended body? 5. Does the escape speed of a rocket depend on its ass? Eplain. 6. Copare the energies required to reach the Moon for a 0 5 -kg spacecraft and a 0 3 -kg satellite. 7. Eplain why it takes ore fuel for a spacecraft to travel fro the Earth to the Moon than for the return trip. Estiate the difference. 8. Why don t we put a geosynchronous weather satellite in orbit around the 45th parallel? Wouldn t this be ore useful for the United States than such a satellite in orbit around the equator? 9. Is the potential energy associated with the Earth Moon syste greater than, less than, or equal to the kinetic energy of the Moon relative to the Earth? 0. Eplain why no work is done on a planet as it oves in a circular orbit around the Sun, even though a gravita- PRBLEMS, 2, 3 = straightforward, interediate, challenging = full solution available in the Student s Manual and Study Guide WEB = solution posted at = Coputer useful in solving proble = Interactive Physics = paired nuerical/sybolic probles Section 4. Newton s Law of Universal Gravitation Section 4.2 Measuring the Gravitational Constant Section 4.3 Free-Fall Acceleration and the Gravitational Force. Deterine the order of agnitude of the gravitational force that you eert on another person 2 away. In your solution, state the quantities that you easure or estiate and their values. 2. A 200-kg ass and a 500-kg ass are separated by (a) Find the net gravitational force eerted by these asses on a 50.0-kg ass placed idway between the. (b) At what position (other than infinitely re- tional force is acting on the planet. What is the net work done on a planet during each revolution as it oves around the Sun in an elliptical orbit?. Eplain why the force eerted on a particle by a unifor sphere ust be directed toward the center of the sphere. Would this be the case if the ass distribution of the sphere were not spherically syetric? 2. Neglecting the density variation of the Earth, what would be the period of a particle oving in a sooth hole dug between opposite points on the Earth s surface, passing through its center? 3. At what position in its elliptical orbit is the speed of a planet a aiu? At what position is the speed a iniu? 4. If you were given the ass and radius of planet X, how would you calculate the free-fall acceleration on the surface of this planet? 5. If a hole could be dug to the center of the Earth, do you think that the force on a ass would still obey Equation 4. there? What do you think the force on would be at the center of the Earth? 6. In his 798 eperient, Cavendish was said to have weighed the Earth. Eplain this stateent. 7. The gravitational force eerted on the Voyager spacecraft by Jupiter accelerated it toward escape speed fro the Sun. How is this possible? 8. How would you find the ass of the Moon? 9. The Apollo 3 spaceship developed trouble in the oygen syste about halfway to the Moon. Why did the spaceship continue on around the Moon and then return hoe, rather than iediately turn back to Earth? ote ones) can the 50.0-kg ass be placed so as to eperience a net force of zero? 3. Three equal asses are located at three corners of a square of edge length, as shown in Figure P4.3. Find the gravitational field g at the fourth corner due to these asses. 4. Two objects attract each other with a gravitational force of agnitude N when separated by 20.0 c. If the total ass of the two objects is 5.00 kg, what is the ass of each? 5. Three unifor spheres of asses 2.00 kg, 4.00 kg, and 6.00 kg are placed at the corners of a right triangle, as illustrated in Figure P4.5. Calculate the resultant gravi-

104 Probles CHAPTER 4 The Law of Gravity WEB y ( 4.00, 0) 6.00 kg tational force on the 4.00-kg ass, assuing that the spheres are isolated fro the rest of the Universe. 6. The free-fall acceleration on the surface of the Moon is about one-sith that on the surface of the Earth. If the radius of the Moon is about 0.250R E, find the ratio of their average densities, Moon / Earth. 7. During a solar eclipse, the Moon, Earth, and Sun all lie on the sae line, with the Moon between the Earth and the Sun. (a) What force is eerted by the Sun on the Moon? (b) What force is eerted by the Earth on the Moon? (c) What force is eerted by the Sun on the Earth? 8. The center-to-center distance between the Earth and the Moon is k. The Moon copletes an orbit in 27.3 days. (a) Deterine the Moon s orbital speed. (b) If gravity were switched off, the Moon would ove along a straight line tangent to its orbit, as described by Newton s first law. In its actual orbit in.00 s, how far does the Moon fall below the tangent line and toward the Earth? 9. When a falling eteoroid is at a distance above the Earth s surface of 3.00 ties the Earth s radius, what is its acceleration due to the Earth s gravity? 0. Two ocean liners, each with a ass of etric tons, are oving on parallel courses, 00 apart. What is the agnitude of the acceleration of one of the liners toward the other due to their utual gravitational attraction? (Treat the ships as point asses.) Figure P4.3 (0, 3.00) F 64 Figure P4.5 y 2.00 kg F kg. A student proposes to easure the gravitational constant G by suspending two spherical asses fro the ceiling of a tall cathedral and easuring the deflection of the cables fro the vertical. Draw a free-body diagra of one of the asses. If two 00.0-kg asses are suspended at the end of long cables, and the cables are attached to the ceiling.000 apart, what is the separation of the asses? 2. n the way to the Moon, the Apollo astronauts reached a point where the Moon s gravitational pull becae stronger than the Earth s. (a) Deterine the distance of this point fro the center of the Earth. (b) What is the acceleration due to the Earth s gravity at this point? Section 4.4 Kepler s Laws Section 4.5 The Law of Gravity and the Motion of Planets 3. A particle of ass oves along a straight line with constant speed in the direction, a distance b fro the ais (Fig. P4.3). Show that Kepler s second law is satisfied by deonstrating that the two shaded triangles in the figure have the sae area when t 4 t 3 t 2 t. v 0 b y t t 2 t 3 t 4 Figure P A counications satellite in geosynchronous orbit reains above a single point on the Earth s equator as the planet rotates on its ais. (a) Calculate the radius of its orbit. (b) The satellite relays a radio signal fro a transitter near the north pole to a receiver, also near the north pole. Traveling at the speed of light, how long is the radio wave in transit? 5. Plaskett s binary syste consists of two stars that revolve in a circular orbit about a center of ass idway between the. This eans that the asses of the two stars are equal (Fig. P4.5). If the orbital velocity of each star is 220 k/s and the orbital period of each is 4.4 days, find the ass M of each star. (For coparison, the ass of our Sun is kg.) 6. Plaskett s binary syste consists of two stars that revolve in a circular orbit about a center of gravity idway between the. This eans that the asses of the two stars are equal (see Fig. P4.5). If the orbital speed of each star is v and the orbital period of each is T, find the ass M of each star. WEB 7. The Eplorer VIII satellite, placed into orbit Noveber 3, 960, to investigate the ionosphere, had the following orbit paraeters: perigee, 459 k; apogee, k (both distances above the Earth s surface); and period, 2.7 in. Find the ratio v p /v a of the speed at perigee to that at apogee. 8. Coet Halley (Fig. P4.8) approaches the Sun to within AU, and its orbital period is 75.6 years (AU is the sybol for astronoical unit, where AU.50 0 is the ean Earth Sun distance). How far fro the Sun will Halley s coet travel before it starts its return journey? AU M Figure P4.5 Probles 5 and 6. Sun 220 k/s 220 k/s 9. Io, a satellite of Jupiter, has an orbital period of.77 days and an orbital radius of k. Fro these data, deterine the ass of Jupiter. 2a CM Figure P4.8 M 20. Two planets, X and Y, travel counterclockwise in circular orbits about a star, as shown in Figure P4.20. The radii of their orbits are in the ratio 3:. At soe tie, they are aligned as in Figure P4.20a, aking a straight line with the star. During the net five years, the angular displaceent of planet X is 90.0, as shown in Figure P4.20b. Where is planet Y at this tie? (a) 2. A synchronous satellite, which always reains above the sae point on a planet s equator, is put in orbit around Jupiter so that scientists can study the faous red spot. Jupiter rotates once every 9.84 h. Use the data in Table 4.2 to find the altitude of the satellite. 22. Neutron stars are etreely dense objects that are fored fro the renants of supernova eplosions. Many rotate very rapidly. Suppose that the ass of a certain spherical neutron star is twice the ass of the Sun and that its radius is 0.0 k. Deterine the greatest possible angular speed it can have for the atter at the surface of the star on its equator to be just held in orbit by the gravitational force. 23. The Solar and Heliospheric bservatory (SH) spacecraft has a special orbit, chosen so that its view of the Sun is never eclipsed and it is always close enough to the Earth to transit data easily. It oves in a nearcircle around the Sun that is saller than the Earth s circular orbit. Its period, however, is not less than yr but is just equal to yr. It is always located between the Earth and the Sun along the line joining the. Both objects eert gravitational forces on the observatory. Show that the spacecraft s distance fro the Earth ust be between and In 772 Joseph Louis Lagrange deterined theoretically the special location that allows this orbit. The SH spacecraft took this position on February 4, 996. (Hint: Use data that are precise to four digits. The ass of the Earth is kg.) Section 4.6 Y X Figure P4.20 The Gravitational Field 24. A spacecraft in the shape of a long cylinder has a length of 00, and its ass with occupants is 000 kg. It has Y X (b)

105 Probles CHAPTER 4 The Law of Gravity k Black hole strayed too close to a.0--radius black hole having a ass 00 ties that of the Sun (Fig. P4.24). The nose of the spacecraft is pointing toward the center of the black hole, and the distance between the nose and the black hole is 0.0 k. (a) Deterine the total force on the spacecraft. (b) What is the difference in the gravitational fields acting on the occupants in the nose of the ship and on those in the rear of the ship, farthest fro the black hole? 25. Copute the agnitude and direction of the gravitational field at a point P on the perpendicular bisector of two equal asses separated by a distance 2a, as shown in Figure P4.25. a M M Figure P4.24 Figure P Find the gravitational field at a distance r along the ais of a thin ring of ass M and radius a. Section 4.7 Gravitational Potential Energy Note: Assue that U 0 as r :. 27. A satellite of the Earth has a ass of 00 kg and is at an altitude of (a) What is the potential energy of the satellite Earth syste? (b) What is the agnitude of the gravitational force eerted by the Earth on the satellite? (c) What force does the satellite eert on the Earth? 28. How uch energy is required to ove a 000-kg ass fro the Earth s surface to an altitude twice the Earth s radius? 29. After our Sun ehausts its nuclear fuel, its ultiate fate ay be to collapse to a white-dwarf state, in which it has approiately the sae ass it has now but a radius r P WEB equal to the radius of the Earth. Calculate (a) the average density of the white dwarf, (b) the acceleration due to gravity at its surface, and (c) the gravitational potential energy associated with a.00-kg object at its surface. 30. At the Earth s surface a projectile is launched straight up at a speed of 0.0 k/s. To what height will it rise? Ignore air resistance. 3. A syste consists of three particles, each of ass 5.00 g, located at the corners of an equilateral triangle with sides of 30.0 c. (a) Calculate the potential energy of the syste. (b) If the particles are released siultaneously, where will they collide? 32. How uch work is done by the Moon s gravitational field as a 000-kg eteor coes in fro outer space and ipacts the Moon s surface? Section 4.8 Energy Considerations in Planetary and Satellite Motion 33. A 500-kg satellite is in a circular orbit at an altitude of 500 k above the Earth s surface. Because of air friction, the satellite is eventually brought to the Earth s surface, and it hits the Earth with a speed of 2.00 k/s. How uch energy was transfored to internal energy by eans of friction? 34. (a) What is the iniu speed, relative to the Sun, that is necessary for a spacecraft to escape the Solar Syste if it starts at the Earth s orbit? (b) Voyager achieved a aiu speed of k/h on its way to photograph Jupiter. Beyond what distance fro the Sun is this speed sufficient for a spacecraft to escape the Solar Syste? 35. A satellite with a ass of 200 kg is placed in Earth orbit at a height of 200 k above the surface. (a) Assuing a circular orbit, how long does the satellite take to coplete one orbit? (b) What is the satellite s speed? (c) What is the iniu energy necessary to place this satellite in orbit (assuing no air friction)? 36. A satellite of ass is placed in Earth orbit at an altitude h. (a) Assuing a circular orbit, how long does the satellite take to coplete one orbit? (b) What is the satellite s speed? (c) What is the iniu energy necessary to place this satellite in orbit (assuing no air friction)? 37. A spaceship is fired fro the Earth s surface with an initial speed of /s. What will its speed be when it is very far fro the Earth? (Neglect friction.) 38. A 000-kg satellite orbits the Earth at a constant altitude of 00 k. How uch energy ust be added to the syste to ove the satellite into a circular orbit at an altitude of 200 k? 39. A treetop satellite oves in a circular orbit just above the surface of a planet, which is assued to offer no air resistance. Show that its orbital speed v and the escape speed fro the planet are related by the epression v esc 2v. 40. The planet Uranus has a ass about 4 ties the Earth s ass, and its radius is equal to about 3.7 Earth radii. (a) By setting up ratios with the corresponding Earth values, find the acceleration due to gravity at the cloud tops of Uranus. (b) Ignoring the rotation of the planet, find the iniu escape speed fro Uranus. 4. Deterine the escape velocity for a rocket on the far side of Ganyede, the largest of Jupiter s oons. The radius of Ganyede is , and its ass is kg. The ass of Jupiter is kg, and the distance between Jupiter and Ganyede is Be sure to include the gravitational effect due to Jupiter, but you ay ignore the otions of Jupiter and Ganyede as they revolve about their center of ass (Fig. P4.4). Jupiter Figure P In Robert Heinlein s The Moon is a Harsh Mistress, the colonial inhabitants of the Moon threaten to launch rocks down onto the Earth if they are not given independence (or at least representation). Assuing that a rail gun could launch a rock of ass at twice the lunar escape speed, calculate the speed of the rock as it enters the Earth s atosphere. (By lunar escape speed we ean the speed required to escape entirely fro a stationary Moon alone in the Universe.) 43. Derive an epression for the work required to ove an Earth satellite of ass fro a circular orbit of radius 2R E to one of radius 3R E. (ptional) Section 4.9 The Gravitational Force Between an Etended bject and a Particle 44. Consider two identical unifor rods of length L and ass lying along the sae line and having their closest points separated by a distance d (Fig. P4.44). Show that the utual gravitational force between these rods has a agnitude F G2 L 2 ln (L d)2 d(2l d) Ganyede 45. A unifor rod of ass M is in the shape of a seicircle of radius R (Fig. P4.45). Calculate the force on a point ass placed at the center of the seicircle. v L (ptional) Section 4.0 The Gravitational Force Between a Particle and a Spherical Mass 46. (a) Show that the period calculated in Eaple 4.0 can be written as d T 2 R E g where g is the free-fall acceleration on the surface of the Earth. (b) What would this period be if tunnels were ade through the Moon? (c) What practical proble regarding these tunnels on Earth would be reoved if they were built on the Moon? 47. A 500-kg unifor solid sphere has a radius of Find the agnitude of the gravitational force eerted by the sphere on a 50.0-g particle located (a).50 fro the center of the sphere, (b) at the surface of the sphere, and (c) fro the center of the sphere. 48. A unifor solid sphere of ass and radius R is inside and concentric with a spherical shell of ass 2 and radius R 2 (Fig. P4.48). Find the gravitational force eerted by the spheres on a particle of ass located at (a) r a, (b) r b, and (c) r c, where r is easured fro the center of the spheres. R 2 Figure P4.44 R Figure P R a M b Figure P4.48 c L

106 Probles CHAPTER 4 The Law of Gravity ADDITINAL PRBLEMS 49. Let g M represent the difference in the gravitational fields produced by the Moon at the points on the Earth s surface nearest to and farthest fro the Moon. Find the fraction g M /g, where g is the Earth s gravitational field. (This difference is responsible for the occurrence of the lunar tides on the Earth.) 50. Two spheres having asses M and 2M and radii R and 3R, respectively, are released fro rest when the distance between their centers is 2R. How fast will each sphere be oving when they collide? Assue that the two spheres interact only with each other. 5. In Larry Niven s science-fiction novel Ringworld, a rigid ring of aterial rotates about a star (Fig. P4.5). The rotational speed of the ring is /s, and its radius is (a) Show that the centripetal acceleration of the inhabitants is 0.2 /s 2. (b) The inhabitants of this ring world eperience a noral contact force n. Acting alone, this noral force would produce an inward acceleration of 9.90 /s 2. Additionally, the star at the center of the ring eerts a gravitational force on the ring and its inhabitants. The difference between the total acceleration and the acceleration provided by the noral force is due to the gravitational attraction of the central star. Show that the ass of the star is approiately 0 32 kg. Star 52. (a) Show that the rate of change of the free-fall acceleration with distance above the Earth s surface is dg dr 2GM E R 3 E This rate of change over distance is called a gradient. (b) If h is sall copared to the radius of the Earth, show that the difference in free-fall acceleration between two points separated by vertical distance h is g 2GM Eh R 3 E n Figure P4.5 F g WEB (c) Evaluate this difference for h 6.00, a typical height for a two-story building. 53. A particle of ass is located inside a unifor solid sphere of radius R and ass M, at a distance r fro its center. (a) Show that the gravitational potential energy of the syste is U (GM/2R 3 )r 2 3GM/2R. (b) Write an epression for the aount of work done by the gravitational force in bringing the particle fro the surface of the sphere to its center. 54. Voyagers and 2 surveyed the surface of Jupiter s oon Io and photographed active volcanoes spewing liquid sulfur to heights of 70 k above the surface of this oon. Find the speed with which the liquid sulfur left the volcano. Io s ass is kg, and its radius is 820 k. 55. As an astronaut, you observe a sall planet to be spherical. After landing on the planet, you set off, walking always straight ahead, and find yourself returning to your spacecraft fro the opposite side after copleting a lap of 25.0 k. You hold a haer and a falcon feather at a height of.40, release the, and observe that they fall together to the surface in 29.2 s. Deterine the ass of the planet. 56. A cylindrical habitat in space, 6.00 k in diaeter and 30 k long, was proposed by G. K. Neill in 974. Such a habitat would have cities, land, and lakes on the inside surface and air and clouds in the center. All of these would be held in place by the rotation of the cylinder about its long ais. How fast would the cylinder have to rotate to iitate the Earth s gravitational field at the walls of the cylinder? 57. In introductory physics laboratories, a typical Cavendish balance for easuring the gravitational constant G uses lead spheres with asses of.50 kg and 5.0 g whose centers are separated by about 4.50 c. Calculate the gravitational force between these spheres, treating each as a point ass located at the center of the sphere. 58. Newton s law of universal gravitation is valid for distances covering an enorous range, but it is thought to fail for very sall distances, where the structure of space itself is uncertain. The crossover distance, far less than the diaeter of an atoic nucleus, is called the Planck length. It is deterined by a cobination of the constants G, c, and h, where c is the speed of light in vacuu and h is Planck s constant (introduced briefly in Chapter and discussed in greater detail in Chapter 40) with units of angular oentu. (a) Use diensional analysis to find a cobination of these three universal constants that has units of length. (b) Deterine the order of agnitude of the Planck length. (Hint: You will need to consider noninteger powers of the constants.) 59. Show that the escape speed fro the surface of a planet of unifor density is directly proportional to the radius of the planet. 60. (a) Suppose that the Earth (or another object) has density (r), which can vary with radius but is spherically WEB syetric. Show that at any particular radius r inside the Earth, the gravitational field strength g(r) will increase as r increases, if and only if the density there eceeds 2/3 the average density of the portion of the Earth inside the radius r. (b) The Earth as a whole has an average density of 5.5 g/c 3, while the density at the surface is.0 g/c 3 on the oceans and about 3 g/c 3 on land. What can you infer fro this? 6. Two hypothetical planets of asses and 2 and radii r and r 2, respectively, are nearly at rest when they are an infinite distance apart. Because of their gravitational attraction, they head toward each other on a collision course. (a) When their center-to-center separation is d, find epressions for the speed of each planet and their relative velocity. (b) Find the kinetic energy of each planet just before they collide, if kg, kg, r , and r (Hint: Both energy and oentu are conserved.) 62. The aiu distance fro the Earth to the Sun (at our aphelion) is.52 0, and the distance of closest approach (at perihelion) is If the Earth s orbital speed at perihelion is k/s, deterine (a) the Earth s orbital speed at aphelion, (b) the kinetic and potential energies at perihelion, and (c) the kinetic and potential energies at aphelion. Is the total energy constant? (Neglect the effect of the Moon and other planets.) 63. A sphere of ass M and radius R has a nonunifor density that varies with r, the distance fro its center, according to the epression Ar, for 0 r R. (a) What is the constant A in ters of M and R? (b) Deterine an epression for the force eerted on a particle of ass placed outside the sphere. (c) Deterine an epression for the force eerted on the particle if it is inside the sphere. (Hint: See Section 4.0 and note that the distribution is spherically syetric.) 64. (a) Deterine the aount of work (in joules) that ust be done on a 00-kg payload to elevate it to a height of 000 k above the Earth s surface. (b) Deterine the aount of additional work that is required to put the payload into circular orbit at this elevation. 65. X-ray pulses fro Cygnus X-, a celestial -ray source, have been recorded during high-altitude rocket flights. The signals can be interpreted as originating when a blob of ionized atter orbits a black hole with a period of 5.0 s. If the blob is in a circular orbit about a black hole whose ass is 20M Sun, what is the orbital radius? 66. Studies of the relationship of the Sun to its galay the Milky Way have revealed that the Sun is located near the outer edge of the galactic disk, about lightyears fro the center. Furtherore, it has been found that the Sun has an orbital speed of approiately 250 k/s around the galactic center. (a) What is the period of the Sun s galactic otion? (b) What is the order of agnitude of the ass of the Milky Way galay? Suppose that the galay is ade ostly of stars, of which the Sun is typical. What is the order of agnitude of the nuber of stars in the Milky Way? 67. The oldest artificial satellite in orbit is Vanguard I, launched March 3, 958. Its ass is.60 kg. In its initial orbit, its iniu distance fro the center of the Earth was 7.02 M, and its speed at this perigee point was 8.23 k/s. (a) Find its total energy. (b) Find the agnitude of its angular oentu. (c) Find its speed at apogee and its aiu (apogee) distance fro the center of the Earth. (d) Find the seiajor ais of its orbit. (e) Deterine its period. 68. A rocket is given an initial speed vertically upward of v i 2 Rg at the surface of the Earth, which has radius R and surface free-fall acceleration g. The rocket otors are quickly cut off, and thereafter the rocket coasts under the action of gravitational forces only. (Ignore atospheric friction and the Earth s rotation.) Derive an epression for the subsequent speed v as a function of the distance r fro the center of the Earth in ters of g, R, and r. 69. Two stars of asses M and, separated by a distance d, revolve in circular orbits about their center of ass (Fig. P4.69). Show that each star has a period given by T d 3 G(M ) (Hint: Apply Newton s second law to each star, and note that the center-of-ass condition requires that Mr 2 r, where r r 2 d.) v r d r 2 Figure P4.69 CM v (a) A 5.00-kg ass is released fro the center of the Earth. It oves with what acceleration relative to the Earth? (b) A kg ass is released fro the center of the Earth. It oves with what acceleration relative to the Earth? Assue that the objects behave as pairs of particles, isolated fro the rest of the Universe. 7. The acceleration of an object oving in the gravitational field of the Earth is a GM E r 3 r M

107 Answers to Quick Quizzes 457 where r is the position vector directed fro the center of the Earth to the object. Choosing the origin at the center of the Earth and assuing that the sall object is oving in the y plane, we find that the rectangular (cartesian) coponents of its acceleration are Use a coputer to set up and carry out a nuerical prea GM E GM ( 2 y 2 ) 3/2 a y E y ( 2 y 2 ) 3/2 diction of the otion of the object, according to Euler s ethod. Assue that the initial position of the object is 0 and y 2R E, where R E is the radius of the Earth. Give the object an initial velocity of /s in the direction. The tie increent should be ade as sall as practical. Try 5 s. Plot the and y coordinates of the object as tie goes on. Does the object hit the Earth? Vary the initial velocity until you find a circular orbit. ANSWERS T QUICK QUIZZES 4. Kepler s third law (Eq. 4.7), which applies to all the planets, tells us that the period of a planet is proportional to r 3/2. Because Saturn and Jupiter are farther fro the Sun than the Earth is, they have longer periods. The Sun s gravitational field is uch weaker at Saturn and Jupiter than it is at the Earth. Thus, these planets eperience uch less centripetal acceleration than the Earth does, and they have correspondingly longer periods. 4.2 The ass of the asteroid ight be so sall that you would be able to eceed escape velocity by leg power alone. You would jup up, but you would never coe back down! 4.3 Kepler s first law applies not only to planets orbiting the Sun but also to any relatively sall object orbiting another under the influence of gravity. Any elliptical path that does not touch the Earth before reaching point G will continue around the other side to point V in a coplete orbit (see figure in net colun). 4.4 The gravitational force is zero inside the shell (Eq. 4.25b). Because the force on it is zero, the particle oves with constant velocity in the direction of its original otion outside the shell until it hits the wall opposite the entry hole. Its path thereafter depends on the nature of the collision and on the particle s original direction.

108 P U Z Z L E R Have you ever wondered why a tennis ball is fuzzy and why a golf ball has diples? A spitball is an illegal baseball pitch because it akes the ball act too uch like the fuzzy tennis ball or the dipled golf ball. What principles of physics govern the behavior of these three pieces of sporting equipent (and also keep airplanes in the sky)? (George Seple) Matter is norally classified as being in one of three states: solid, liquid, or gas. Fro everyday eperience, we know that a solid has a definite volue and shape. A brick aintains its failiar shape and size day in and day out. We also know that a liquid has a definite volue but no definite shape. Finally, we know that an unconfined gas has neither a definite volue nor a definite shape. These definitions help us picture the states of atter, but they are soewhat artificial. For eaple, asphalt and plastics are norally considered solids, but over long periods of tie they tend to flow like liquids. Likewise, ost substances can be a solid, a liquid, or a gas (or a cobination of any of these), depending on the teperature and pressure. In general, the tie it takes a particular substance to change its shape in response to an eternal force deterines whether we treat the substance as a solid, as a liquid, or as a gas. A fluid is a collection of olecules that are randoly arranged and held together by weak cohesive forces and by forces eerted by the walls of a container. Both liquids and gases are fluids. In our treatent of the echanics of fluids, we shall see that we do not need to learn any new physical principles to eplain such effects as the buoyant force acting on a suberged object and the dynaic lift acting on an airplane wing. First, we consider the echanics of a fluid at rest that is, fluid statics and derive an epression for the pressure eerted by a fluid as a function of its density and depth. We then treat the echanics of fluids in otion that is, fluid dynaics. We can describe a fluid in otion by using a odel in which we ake certain siplifying assuptions. We use this odel to analyze soe situations of practical iportance. An analysis leading to Bernoulli s equation enables us to deterine relationships between the pressure, density, and velocity at every point in a fluid. 5. Pressure 459 c h a p t e r Fluid Mechanics Chapter utline 5. Pressure 5.2 Variation of Pressure with Depth 5.3 Pressure Measureents 5.4 Buoyant Forces and Archiedes s Principle 5.5 Fluid Dynaics 5.6 Strealines and the Equation of Continuity 5.7 Bernoulli s Equation 5.8 (ptional) ther Applications of Bernoulli s Equation 5. PRESSURE Fluids do not sustain shearing stresses or tensile stresses; thus, the only stress that can be eerted on an object suberged in a fluid is one that tends to copress the object. In other words, the force eerted by a fluid on an object is always perpendicular to the surfaces of the object, as shown in Figure 5.. The pressure in a fluid can be easured with the device pictured in Figure 5.2. The device consists of an evacuated cylinder that encloses a light piston connected to a spring. As the device is suberged in a fluid, the fluid presses on the top of the piston and copresses the spring until the inward force eerted by the fluid is balanced by the outward force eerted by the spring. The fluid pressure can be easured directly if the spring is calibrated in advance. If F is the agnitude of the force eerted on the piston and A is the surface area of the piston, Vacuu A F Figure 5.2 A siple device for easuring the pressure eerted by a fluid. Figure 5. At any point on the surface of a suberged object, the force eerted by the fluid is perpendicular to the surface of the object. The force eerted by the fluid on the walls of the container is perpendicular to the walls at all points. 458

109 460 CHAPTER 5 Fluid Mechanics 5.2 Variation of Pressure with Depth 46 Definition of pressure Snowshoes keep you fro sinking into soft snow because they spread the downward force you eert on the snow over a large area, reducing the pressure on the snow s surface. QuickLab Place a tack between your thub and inde finger, as shown in the figure. Now very gently squeeze the tack and note the sensation. The pointed end of the tack causes pain, and the blunt end does not. According to Newton s third law, the force eerted by the tack on the thub is equal in agnitude and opposite in direction to the force eerted by the tack on the inde finger. However, the pressure at the pointed end of the tack is uch greater than the pressure at the blunt end. (Reeber that pressure is force per unit area.) Tack then the pressure P of the fluid at the level to which the device has been suberged is defined as the ratio F/A: P F (5.) A Note that pressure is a scalar quantity because it is proportional to the agnitude of the force on the piston. To define the pressure at a specific point, we consider a fluid acting on the device shown in Figure 5.2. If the force eerted by the fluid over an infinitesial surface eleent of area da containing the point in question is df, then the pressure at that point is P df (5.2) da As we shall see in the net section, the pressure eerted by a fluid varies with depth. Therefore, to calculate the total force eerted on a flat wall of a container, we ust integrate Equation 5.2 over the surface area of the wall. Because pressure is force per unit area, it has units of newtons per square eter (N/ 2 ) in the SI syste. Another nae for the SI unit of pressure is pascal (Pa): Pa N/ 2 (5.3) Quick Quiz 5. Suppose you are standing directly behind soeone who steps back and accidentally stops on your foot with the heel of one shoe. Would you be better off if that person were a professional basketball player wearing sneakers or a petite woan wearing spike-heeled shoes? Eplain. Quick Quiz 5.2 After a long lecture, the daring physics professor stretches out for a nap on a bed of nails, as shown in Figure 5.3. How is this possible? Figure 5.3 EXAMPLE VARIATIN F PRESSURE WITH DEPTH As divers well know, water pressure increases with depth. Likewise, atospheric pressure decreases with increasing altitude; it is for this reason that aircraft flying at high altitudes ust have pressurized cabins. We now show how the pressure in a liquid increases linearly with depth. As Equation. describes, the density of a substance is defined as its ass per unit volue: Table 5. lists the densities of various substances. These values vary slightly with teperature because the volue of a substance is teperature dependent (as we shall see in Chapter 9). Note that under standard conditions (at 0 C and at atospheric pressure) the densities of gases are about / 000 the densities of solids and liquids. This difference iplies that the average olecular spacing in a gas under these conditions is about ten ties greater than that in a solid or liquid. Now let us consider a fluid of density at rest and open to the atosphere, as shown in Figure 5.4. We assue that is constant; this eans that the fluid is incopressible. Let us select a saple of the liquid contained within an iaginary cylinder of cross-sectional area A etending fro the surface to a depth h. The /V. The Water Bed The attress of a water bed is 2.00 long by 2.00 wide and 30.0 c deep. (a) Find the weight of the water in the attress. The density of water is 000 kg/ 3 (Table 5.), and so the ass of the water is M V ( 000 kg/ 3 )(.20 3 ) kg and its weight is Mg ( kg)(9.80 /s 2 ) This is approiately lb. (A regular bed weighs appro- TABLE N iately 300 lb.) Because this load is so great, such a water bed is best placed in the baseent or on a sturdy, wellsupported floor. (b) Find the pressure eerted by the water on the floor when the bed rests in its noral position. Assue that the entire lower surface of the bed akes contact with the floor. Densities of Soe Coon Substances at Standard Teperature (0 C) and Pressure (Atospheric) Substance (kg/ 3 ) Substance (kg/ 3 ) Air.29 Ice Aluinu Iron Benzene Lead Copper Mercury Ethyl alcohol ak Fresh water ygen gas.43 Glycerine Pine Gold Platinu Heliu gas.79 0 Seawater Hydrogen gas Silver When the bed is in its noral position, the crosssectional area is ; thus, fro Equation 5., we find that P.8 04 N Pa h Mg P 0 Aj PAj Figure 5.4 How pressure varies with depth in a fluid. The net force eerted on the volue of water within the darker region ust be zero.

110 462 CHAPTER 5 Fluid Mechanics 5.2 Variation of Pressure with Depth 463 A QuickLab Poke two holes in the side of a paper or polystyrene cup one near the top and the other near the botto. Fill the cup with water and watch the water flow out of the holes. Why does water eit fro the botto hole at a higher speed than it does fro the top hole? Variation of pressure with depth B C This arrangeent of interconnected tubes deonstrates that the pressure in a liquid is the sae at all points having the sae elevation. For eaple, the pressure is the sae at points A, B, C, and D. D pressure eerted by the outside liquid on the botto face of the cylinder is P, and the pressure eerted on the top face of the cylinder is the atospheric pressure P 0. Therefore, the upward force eerted by the outside fluid on the botto of the cylinder is PA, and the downward force eerted by the atosphere on the top is P 0 A. The ass of liquid in the cylinder is M V Ah; therefore, the weight of the liquid in the cylinder is Mg Ahg. Because the cylinder is in equilibriu, the net force acting on it ust be zero. Choosing upward to be the positive y direction, we see that F y PA P 0 A Mg 0 or PA P 0 A Ahg 0 PA P 0 A Ahg P P 0 gh (5.4) That is, the pressure P at a depth h below the surface of a liquid open to the atosphere is greater than atospheric pressure by an aount gh. In our calculations and working of end-of-chapter probles, we usually take atospheric pressure to be Equation 5.4 iplies that the pressure is the sae at all points having the sae depth, independent of the shape of the container. Quick Quiz 5.3 In the derivation of Equation 5.4, why were we able to ignore the pressure that the liquid eerts on the sides of the cylinder? In view of the fact that the pressure in a fluid depends on depth and on the value of P 0, any increase in pressure at the surface ust be transitted to every other point in the fluid. This concept was first recognized by the French scientist Blaise Pascal ( ) and is called Pascal s law: A change in the pressure applied to a fluid is transitted undiinished to every point of the fluid and to the walls of the container. An iportant application of Pascal s law is the hydraulic press illustrated in Figure 5.5a. A force of agnitude F is applied to a sall piston of surface area A. The pressure is transitted through a liquid to a larger piston of surface area A 2. Because the pressure ust be the sae on both sides, P F /A F 2 /A 2. Therefore, the force F 2 is greater than the force F by a factor A 2 /A, which is called the force-ultiplying factor. Because liquid is neither added nor reoved, the volue pushed down on the left as the piston oves down a distance d equals the volue pushed up on the right as the right piston oves up a distance d 2. That is, A d A 2 d 2 ; thus, the force-ultiplying factor can also be written as d /d 2. Note that F d F 2 d 2. Hydraulic brakes, car lifts, hydraulic jacks, and forklifts all ake use of this principle (Fig. 5.5b). Quick Quiz 5.4 P 0.00 at Pa A grain silo has any bands wrapped around its perieter (Fig. 5.6). Why is the spacing between successive bands saller at the lower portions of the silo, as shown in the photograph? F d A 2 A d 2 EXAMPLE 5.2 The Car Lift In a car lift used in a service station, copressed air eerts a force on a sall piston that has a circular cross section and a radius of 5.00 c. This pressure is transitted by a liquid to a piston that has a radius of 5.0 c. What force ust the copressed air eert to lift a car weighing N? What air pressure produces this force? (a) F 2 Figure 5.5 (a) Diagra of a hydraulic press. Because the increase in pressure is the sae on the two sides, a sall force F l at the left produces a uch greater force F 2 at the right. (b) A vehicle undergoing repair is supported by a hydraulic lift in a garage. Because the pressure eerted by the copressed air is transitted undiinished throughout the liquid, we have F A A 2 F ( ) 2 2 ( ) ( N) N EXAMPLE 5.3 A Pain in the Ear Estiate the force eerted on your eardru due to the water above when you are swiing at the botto of a pool that is 5.0 deep. First, we ust find the unbalanced pressure on (b) Figure 5.6 The air pressure that produces this force is P F N A ( ) Pa This pressure is approiately twice atospheric pressure. The input work (the work done by F ) is equal to the output work (the work done by F 2 ), in accordance with the principle of conservation of energy. the eardru; then, after estiating the eardru s surface area, we can deterine the force that the water eerts on it. The air inside the iddle ear is norally at atospheric pressure P 0. Therefore, to find the net force on the eardru, we ust consider the difference between the total pressure at

111 464 CHAPTER 5 Fluid Mechanics 5.4 Buoyant Forces and Archiedes s Principle 465 the botto of the pool and atospheric pressure: P bot P 0 gh ( kg/ 3 )(9.80 /s 2 )(5.0 ) Pa We estiate the surface area of the eardru to be approiately c This eans that the force on it is F (P bot P 0 )A 5 N. Because a force on the eardru of this agnitude is etreely uncofortable, swiers often pop their ears while under water, an action that pushes air fro the lungs into the iddle ear. Using this technique equalizes the pressure on the two sides of the eardru and relieves the discofort. P A h P 0 B h P 0 P = 0 EXAMPLE 5.4 The Force on a Da Water is filled to a height H behind a da of width w (Fig. 5.7). Deterine the resultant force eerted by the water on the da. Because pressure varies with depth, we cannot calculate the force siply by ultiplying the area by the pressure. We can solve the proble by finding the force df e- Figure 5.7 H h w Because pressure varies with depth, the total force eerted on a da ust be obtained fro the epression F P da, where da is the area of the dark strip. dy y 5.3 PRESSURE erted on a narrow horizontal strip at depth h and then integrating the epression to find the total force. Let us iagine a vertical y ais, with y 0 at the botto of the da and our strip a distance y above the botto. We can use Equation 5.4 to calculate the pressure at the depth h; we oit atospheric pressure because it acts on both sides of the da: Using Equation 5.2, we find that the force eerted on the shaded strip of area da w dy is Therefore, the total force on the da is F P da H Note that the thickness of the da shown in Figure 5.7 increases with depth. This design accounts for the greater and greater pressure that the water eerts on the da at greater depths. Eercise Find an epression for the average pressure on the da fro the total force eerted by the water on the da. Answer 2 gh. MEASUREMENTS P gh g(h y) df P da g(h y)w dy 2gwH 2 ne siple device for easuring pressure is the open-tube anoeter illustrated in Figure 5.8a. ne end of a U-shaped tube containing a liquid is open to the atosphere, and the other end is connected to a syste of unknown pressure P. The difference in pressure P P 0 is equal to gh; hence, P P 0 gh. The pressure P is called the absolute pressure, and the difference P P 0 is called the gauge pressure. The latter is the value that norally appears on a pressure gauge. For eaple, the pressure you easure in your bicycle tire is the gauge pressure. Another instruent used to easure pressure is the coon baroeter, which was invented by Evangelista Torricelli ( ). The baroeter consists of a 0 g(h y)w dy long, ercury-filled tube closed at one end and inverted into an open container of ercury (Fig. 5.8b). The closed end of the tube is nearly a vacuu, and so its pressure can be taken as zero. Therefore, it follows that P 0 gh, where h is the height of the ercury colun. ne atosphere (P 0 at) of pressure is defined as the pressure that causes the colun of ercury in a baroeter tube to be eactly in height at 0 C, with g /s 2. At this teperature, ercury has a density of kg/ 3 ; therefore, Quick Quiz 5.5 ther than the obvious proble that occurs with freezing, why don t we use water in a baroeter in the place of ercury? 5.4 (a) Figure 5.8 Two devices for easuring pressure: (a) an open-tube anoeter and (b) a ercury baroeter. P 0 gh ( kg/ 3 )( /s 2 )( ) Pa at BUYANT FRCES AND ARCHIMEDES S PRINCIPLE Have you ever tried to push a beach ball under water? This is etreely difficult to do because of the large upward force eerted by the water on the ball. The upward force eerted by water on any iersed object is called a buoyant force. We can deterine the agnitude of a buoyant force by applying soe logic and Newton s second law. Iagine that, instead of air, the beach ball is filled with water. If you were standing on land, it would be difficult to hold the water-filled ball in your ars. If you held the ball while standing neck deep in a pool, however, the force you would need to hold it would alost disappear. In fact, the required force would be zero if we were to ignore the thin layer of plastic of which the beach ball is ade. Because the water-filled ball is in equilibriu while it is suberged, the agnitude of the upward buoyant force ust equal its weight. If the suberged ball were filled with air rather than water, then the upward buoyant force eerted by the surrounding water would still be present. However, because the weight of the water is now replaced by the uch saller weight of that volue of air, the net force is upward and quite great; as a result, the ball is pushed to the surface. (b)

112 466 CHAPTER 5 Fluid Mechanics 5.4 Buoyant Forces and Archiedes s Principle 467 Archiedes s principle Archiedes (c B.C.) Archiedes, a Greek atheatician, physicist, and engineer, was perhaps the greatest scientist of antiquity. He was the first to copute accurately the ratio of a circle s circuference to its diaeter, and he showed how to calculate the volue and surface area of spheres, cylinders, and other geoetric shapes. He is well known for discovering the nature of the buoyant force. Archiedes was also a gifted inventor. ne of his practical inventions, still in use today, is Archiedes s screw an inclined, rotating, coiled tube originally used to lift water fro the holds of ships. He also invented the catapult and devised systes of levers, pulleys, and weights for raising heavy loads. Such inventions were successfully used to defend his native city Syracuse during a two-year siege by the Roans. The anner in which buoyant forces act is suarized by Archiedes s principle, which states that the agnitude of the buoyant force always equals the weight of the fluid displaced by the object. The buoyant force acts vertically upward through the point that was the center of gravity of the displaced fluid. Note that Archiedes s principle does not refer to the akeup of the object eperiencing the buoyant force. The object s coposition is not a factor in the buoyant force. We can verify this in the following anner: Suppose we focus our attention on the indicated cube of liquid in the container illustrated in Figure 5.9. This cube is in equilibriu as it is acted on by two forces. ne of these forces is the gravitational force F g. What cancels this downward force? Apparently, the rest of the liquid in the container is holding the cube in equilibriu. Thus, the agnitude of the buoyant force B eerted on the cube is eactly equal to the agnitude of F g, which is the weight of the liquid inside the cube: B F g Now iagine that the cube of liquid is replaced by a cube of steel of the sae diensions. What is the buoyant force acting on the steel? The liquid surrounding a cube behaves in the sae way no atter what the cube is ade of. Therefore, the buoyant force acting on the steel cube is the sae as the buoyant force acting on a cube of liquid of the sae diensions. In other words, the agnitude of the buoyant force is the sae as the weight of the liquid cube, not the steel cube. Although atheatically ore coplicated, this sae principle applies to suberged objects of any shape, size, or density. Although we have described the agnitude and direction of the buoyant force, we still do not know its origin. Why would a fluid eert such a strange force, alost as if the fluid were trying to epel a foreign body? To understand why, look again at Figure 5.9. The pressure at the botto of the cube is greater than the pressure at the top by an aount gh, where h is the length of any side of the cube. The pressure difference P between the botto and top faces of the cube is equal to the buoyant force per unit area of those faces that is, P B/A. Therefore, B (P)A (gh)a gv, where V is the volue of the cube. Because the ass of the fluid in the cube is M V, we see that B F g Vg Mg (5.5) where Mg is the weight of the fluid in the cube. Thus, the buoyant force is a result of the pressure differential on a suberged or partly suberged object. Before we proceed with a few eaples, it is instructive for us to copare the forces acting on a totally suberged object with those acting on a floating (partly suberged) object. a (a) B F g case, the upward buoyant force is balanced by the downward gravitational force acting on the object. If V f is the volue of the fluid displaced by the object (this volue is the sae as the volue of that part of the object that is beneath the fluid level), the buoyant force has a agnitude B fv f g. Because the weight of the object is F g Mg ov o g, and because F g B, we see that fv f g ov o g, or o Vf (5.6) f V o Under noral conditions, the average density of a fish is slightly greater than the density of water. It follows that the fish would sink if it did not have soe echanis for adjusting its density. The fish accoplishes this by internally regulating the size of its air-filled swi bladder to balance the change in the agnitude of the buoyant force acting on it. In this anner, fish are able to swi to various depths. Unlike a fish, a scuba diver cannot achieve neutral buoyancy (at which the buoyant force just balances the weight) by adjusting the agnitude of the buoyant force B. Instead, the diver adjusts F g by anipulating lead weights. Quick Quiz 5.6 Steel is uch denser than water. In view of this fact, how do steel ships float? Quick Quiz 5.7 a (b) B F g Figure 5.0 (a) A totally suberged object that is less dense than the fluid in which it is suberged eperiences a net upward force. (b) A totally suberged object that is denser than the fluid sinks. A glass of water contains a single floating ice cube (Fig. 5.). When the ice elts, does the water level go up, go down, or reain the sae? Hot-air balloons. Because hot air is less dense than cold air, a net upward force acts on the balloons. h F g Figure 5.9 The eternal forces acting on the cube of liquid are the force of gravity F g and the buoyant force B. Under equilibriu conditions, B F g. B Case : Totally Suberged bject When an object is totally suberged in a fluid of density f, the agnitude of the upward buoyant force is B fv o g, where V o is the volue of the object. If the object has a ass M and density o, its weight is equal to F g Mg ov o g, and the net force on it is B F g (f o)v o g. Hence, if the density of the object is less than the density of the fluid, then the downward force of gravity is less than the buoyant force, and the unconstrained object accelerates upward (Fig. 5.0a). If the density of the object is greater than the density of the fluid, then the upward buoyant force is less than the downward force of gravity, and the unsupported object sinks (Fig. 5.0b). Case 2: Floating bject Now consider an object of volue V o in static equilibriu floating on a fluid that is, an object that is only partially suberged. In this Quick Quiz 5.8 When a person in a rowboat in a sall pond throws an anchor overboard, does the water level of the pond go up, go down, or reain the sae? EXAMPLE 5.5 Eureka! Archiedes supposedly was asked to deterine whether a crown ade for the king consisted of pure gold. Legend has it that he solved this proble by weighing the crown first in air and then in water, as shown in Figure 5.2. Suppose the Figure 5. scale read 7.84 N in air and 6.86 N in water. What should Archiedes have told the king? When the crown is suspended in air, the scale

113 468 CHAPTER 5 Fluid Mechanics 5.5 Fluid Dynaics 469 reads the true weight T F g (neglecting the buoyancy of air). When it is iersed in water, the buoyant force B reduces the scale reading to an apparent weight of T 2 F g B. Hence, the buoyant force eerted on the crown is the difference between its weight in air and its weight in water: B F g T N 6.86 N 0.98 N he had been cheated. Either the crown was hollow, or it was not ade of pure gold. 5.5 FLUID DYNAMICS Thus far, our study of fluids has been restricted to fluids at rest. We now turn our attention to fluids in otion. Instead of trying to study the otion of each particle of the fluid as a function of tie, we describe the properties of a oving fluid at each point as a function of tie. Because this buoyant force is equal in agnitude to the weight of the displaced water, we have wgv w 0.98 N, where V w is the volue of the displaced water and w is its density. Also, the volue of the crown V c is equal to the volue of the displaced water because the crown is copletely suberged. Therefore, V c V w 0.98 N 0.98 N gw (9.8 /s 2 )( 000 kg/ 3 ) Finally, the density of the crown is c c c g V c V c g 7.84 N ( )(9.8 /s 2 ) kg/ 3 Fro Table 5. we see that the density of gold is kg/ 3. Thus, Archiedes should have told the king that T T 2 F g Figure 5.2 (a) (a) When the crown is suspended in air, the scale reads its true weight T F g (the buoyancy of air is negligible). (b) When the crown is iersed in water, the buoyant force B reduces the scale reading to the apparent weight T 2 F g B. B F g (b) Flow Characteristics When fluid is in otion, its flow can be characterized as being one of two ain types. The flow is said to be steady, or lainar, if each particle of the fluid follows a sooth path, such that the paths of different particles never cross each other, as shown in Figure 5.4. In steady flow, the velocity of the fluid at any point reains constant in tie. Above a certain critical speed, fluid flow becoes turbulent; turbulent flow is irregular flow characterized by sall whirlpool-like regions, as shown in Figure 5.5. The ter viscosity is coonly used in the description of fluid flow to characterize the degree of internal friction in the fluid. This internal friction, or viscous force, is associated with the resistance that two adjacent layers of fluid have to oving relative to each other. Viscosity causes part of the kinetic energy of a fluid to be converted to internal energy. This echanis is siilar to the one by which an object sliding on a rough horizontal surface loses kinetic energy. Because the otion of real fluids is very cople and not fully understood, we ake soe siplifying assuptions in our approach. In our odel of an ideal fluid, we ake the following four assuptions:. The fluid is nonviscous. In a nonviscous fluid, internal friction is neglected. An object oving through the fluid eperiences no viscous force. 2. The flow is steady. In steady (lainar) flow, the velocity of the fluid at each point reains constant. EXAMPLE 5.6 A Titanic Surprise An iceberg floating in seawater, as shown in Figure 5.3a, is etreely dangerous because uch of the ice is below the surface. This hidden ice can daage a ship that is still a considerable distance fro the visible ice. What fraction of the iceberg lies below the water level? This proble corresponds to Case 2. The weight of the iceberg is F where kg/ 3 gi iv i g, i 97 and V i is the volue of the whole iceberg. The agnitude of the up- ward buoyant force equals the weight of the displaced water: B wv w g, where V w, the volue of the displaced water, is equal to the volue of the ice beneath the water (the shaded region in Fig. 5.3b) and w is the density of seawater, w 030 kg/ 3. Because iv i g wv w g, the fraction of ice beneath the water s surface is f V w i 97 kg/3 V i w 030 kg/ or 89.0% Properties of an ideal fluid (a) (b) Figure 5.3 (a) Much of the volue of this iceberg is beneath the water. (b) A ship can be daaged even when it is not near the eposed ice. Figure 5.4 Lainar flow around an autoobile in a test wind tunnel. Figure 5.5 Hot gases fro a cigarette ade visible by soke particles. The soke first oves in lainar flow at the botto and then in turbulent flow above.

114 470 CHAPTER 5 Fluid Mechanics 5.7 Bernoulli s Equation 47 P Figure 5.6 A particle in lainar flow follows a strealine, and at each point along its path the particle s velocity is tangent to the strealine. Equation of continuity v 3. The fluid is incopressible. The density of an incopressible fluid is constant. 4. The flow is irrotational. In irrotational flow, the fluid has no angular oentu about any point. If a sall paddle wheel placed anywhere in the fluid does not rotate about the wheel s center of ass, then the flow is irrotational. 5.6 STREAMLINES AND THE EQUATIN F CNTINUITY The path taken by a fluid particle under steady flow is called a strealine. The velocity of the particle is always tangent to the strealine, as shown in Figure 5.6. A set of strealines like the ones shown in Figure 5.6 for a tube of flow. Note that fluid particles cannot flow into or out of the sides of this tube; if they could, then the strealines would cross each other. Consider an ideal fluid flowing through a pipe of nonunifor size, as illustrated in Figure 5.7. The particles in the fluid ove along strealines in steady flow. In a tie t, the fluid at the botto end of the pipe oves a distance v t. If A is the cross-sectional area in this region, then the ass of fluid contained in the left shaded region in Figure 5.7 is A A v t, where is the (nonchanging) density of the ideal fluid. Siilarly, the fluid that oves through the upper end of the pipe in the tie t has a ass 2 A 2 v 2 t. However, because ass is conserved and because the flow is steady, the ass that crosses A in a tie t ust equal the ass that crosses A 2 in the tie t. That is, 2, or A v t A 2 v 2 t ; this eans that A v A 2 v 2 constant This epression is called the equation of continuity. It states that (5.7) the product of the area and the fluid speed at all points along the pipe is a constant for an incopressible fluid. This equation tells us that the speed is high where the tube is constricted (sall A) and low where the tube is wide (large A). The product Av, which has the diensions of volue per unit tie, is called either the volue flu or the flow rate. The condition Av constant is equivalent to the stateent that the volue of fluid that enters one end of a tube in a given tie interval equals the volue leaving the other end of the tube in the sae tie interval if no leaks are present. A v v 2 2 A 2 Figure 5.7 A fluid oving with steady flow through a pipe of varying cross-sectional area. The volue of fluid flowing through area A in a tie interval t ust equal the volue flowing through area A 2 in the sae tie interval. Therefore, A v A 2 v 2. EXAMPLE BERNULLI S Niagara Falls Each second, of water flows over the 670--wide cliff of the Horseshoe Falls portion of Niagara Falls. The water is approiately 2 deep as it reaches the cliff. What is its speed at that instant? The cross-sectional area of the water as it reaches the edge of the cliff is A (670 )(2 ) The flow rate of /s is equal to Av. This gives v /s A /s EQUATIN 4 /s When you press your thub over the end of a garden hose so that the opening becoes a sall slit, the water coes out at high speed, as shown in Figure 5.9. Is the water under greater pressure when it is inside the hose or when it is out in the air? You can answer this question by noting how hard you have to push your thub against the water inside the end of the hose. The pressure inside the hose is definitely greater than atospheric pressure. The relationship between fluid speed, pressure, and elevation was first derived in 738 by the Swiss physicist Daniel Bernoulli. Consider the flow of an ideal fluid through a nonunifor pipe in a tie t, as illustrated in Figure Let us call the lower shaded part section and the upper shaded part section 2. The force eerted by the fluid in section has a agnitude P A. The work done by this force in a tie t is W F P A P V, where V is the volue of section. In a siilar anner, the work done by the fluid in section 2 in the sae tie t is W 2 P 2 A 2 2 P 2 V. (The volue that passes through section in a tie t equals the volue that passes through section 2 in the sae tie.) This work is negative because the fluid force opposes the displaceent. Thus, the net work done by these forces in the tie t is W (P P 2 )V P A y v Note that we have kept only one significant figure because our value for the depth has only one significant figure. Eercise A barrel floating along in the river plunges over the Falls. How far fro the base of the cliff is the barrel when it reaches the water 49 below? Answer 3 0. y 2 2 v 2 P 2 A2 Daniel Bernoulli ( ) Daniel Bernoulli, a Swiss physicist and atheatician, ade iportant discoveries in fluid dynaics. Born into a faily of atheaticians, he was the only eber of the faily to ake a ark in physics. Bernoulli s ost faous work, Hydrodynaica, was published in 738; it is both a theoretical and a practical study of equilibriu, pressure, and speed in fluids. He showed that as the speed of a fluid increases, its pressure decreases. In Hydrodynaica Bernoulli also attepted the first eplanation of the behavior of gases with changing pressure and teperature; this was the beginning of the kinetic theory of gases, a topic we study in Chapter 2. (Corbis Bettann) Figure 5.8 Quick Quiz 5.9 As water flows fro a faucet, as shown in Figure 5.8, why does the strea of water becoe narrower as it descends? Figure 5.9 The speed of water spraying fro the end of a hose increases as the size of the opening is decreased with the thub. Figure 5.20 A fluid in lainar flow through a constricted pipe. The volue of the shaded section on the left is equal to the volue of the shaded section on the right.

115 472 CHAPTER 5 Fluid Mechanics 5.7 Bernoulli s Equation 473 QuickLab Place two soda cans on their sides approiately 2 c apart on a table. Align your outh at table level and with the space between the cans. Blow a horizontal strea of air through this space. What do the cans do? Is this what you epected? Copare this with the force acting on a car parked close to the edge of a road when a big truck goes by. How does the outcoe relate to Equation 5.9? Bernoulli s equation EXAMPLE 5.8 Part of this work goes into changing the kinetic energy of the fluid, and part goes into changing the gravitational potential energy. If is the ass that enters one end and leaves the other in a tie t, then the change in the kinetic energy of this ass is K 2 v v 2 The change in gravitational potential energy is U g y 2 g y We can apply Equation 8.3, W K U, to this volue of fluid to obtain (P P 2 )V 2 v v 2 g y 2 g y If we divide each ter by V and recall that this epression reduces to P P 2 2v 2 2 2v 2 gy 2 gy Rearranging ters, we obtain P 2v 2 gy P 2 2v 2 2 gy 2 (5.8) This is Bernoulli s equation as applied to an ideal fluid. It is often epressed as (5.9) This epression specifies that, in lainar flow, the su of the pressure (P), kinetic energy per unit volue ( 2v 2 ), and gravitational potential energy per unit volue (gy) has the sae value at all points along a strealine. When the fluid is at rest, v v 2 0 and Equation 5.8 becoes This is in agreeent with Equation 5.4. The Venturi Tube The horizontal constricted pipe illustrated in Figure 5.2, known as a Venturi tube, can be used to easure the flow speed of an incopressible fluid. Let us deterine the flow speed at point 2 if the pressure difference P P 2 is known. Because the pipe is horizontal, y y 2, and applying Equation 5.8 to points and 2 gives () P 2v 2 P 2 2 2v 2 /V, P 2v 2 gy constant P P 2 g(y 2 y ) gh Fro the equation of continuity, A v A 2 v 2, we find that (2) v A 2 v A 2 Substituting this epression into equation () gives P 2 A 2 A 2 v 2 2 P 2 2 2v 2 EXAMPLE 5.9 A Good Trick It is possible to blow a die off a table and into a tubler. Place the die about 2 c fro the edge of the table. Place the tubler on the table horizontally with its open edge about 2 c fro the die, as shown in Figure 5.22a. If you blow forcefully across the top of the die, it will rise, be caught in the airstrea, and end up in the tubler. The 2 c v 2 2 c (a) A! 2(P P 2 ) (A 2 A 2 2 ) We can use this result and the continuity equation to obtain an epression for v. Because A 2 A, Equation (2) shows us that v 2 v. This result, together with equation (), indicates that P P 2. In other words, the pressure is reduced in the constricted part of the pipe. This result is soewhat analogous to the following situation: Consider a very crowded roo in which people are squeezed together. As soon as a door is opened and people begin to eit, the squeezing (pressure) is least near the door, where the otion (flow) is greatest. ass of a die is 2.24 g, and its surface area is A How hard are you blowing when the die rises and travels into the tubler? Figure 5.22b indicates we ust calculate the upward force acting on the die. First, note that a thin stationary layer of air is present between the die and the table. When you blow across the die, it deflects ost of the oving air fro your breath across its top, so that the air above the die has a greater speed than the air beneath it. This fact, together with Bernoulli s equation, deonstrates that the air oving across the top of the die is at a lower pressure than the air beneath the die. If we neglect the sall thickness of the die, we can apply Equation 5.8 to obtain P above 2v 2 above P beneath 2v 2 beneath Because the air beneath the die is alost stationary, we can neglect the last ter in this epression and write the difference as P beneath P above 2v 2 above. If we ultiply this pressure difference by the surface area of the die, we obtain the upward force acting on the die. At the very least, this upward force ust balance the gravitational force acting on the die, and so, taking the density of air fro Table 5., we can state that F g g (P beneath P above )A ( 2v 2 above)a P P2 F Bernoulli v above! 2g A! 2( kg)(9.80 /s2 ) (.29 kg/ 3 )( ) Figure 5.2 (a) Pressure P is greater than pressure P 2 because v v 2. This device can be used to easure the speed of fluid flow. (b) A Venturi tube. A v v 2 (a) A 2 (b) F g (b) Figure 5.22 v above.7 /s The air you blow ust be oving faster than this if the upward force is to eceed the weight of the die. Practice this trick a few ties and then ipress all your friends!

116 474 CHAPTER 5 Fluid Mechanics 5.8 ther Applications of Bernoulli s Equation 475 EXAMPLE 5.0 Torricelli s Law An enclosed tank containing a liquid of density has a hole in its side at a distance y fro the tank s botto (Fig. 5.23). The hole is open to the atosphere, and its diaeter is uch saller than the diaeter of the tank. The air above the liquid is aintained at a pressure P. Deterine the speed at 2 P A 2 which the liquid leaves the hole when the liquid s level is a distance h above the hole. Because A 2 W A, the liquid is approiately at rest at the top of the tank, where the pressure is P. Applying Bernoulli s equation to points and 2 and noting that at the hole P is equal to atospheric pressure P 0, we find that P 0 2v 2 gy P gy 2 But y 2 y h; thus, this epression reduces to QuickLab You can easily deonstrate the effect of changing fluid direction by lightly holding the back of a spoon against a strea of water coing fro a faucet. You will see the strea attach itself to the curvature of the spoon and be deflected sideways. You will also feel the third-law force eerted by the water on the spoon. Figure 5.23 y 2 h y When P is uch larger than atospheric pressure P 0, the liquid speed as the liquid passes through the hole in the side of the container is given approiately by v!2(p P 0 )/. A P 0 v v! 2(P P 0) 2gh When P is uch greater than P 0 (so that the ter 2gh can be neglected), the eit speed of the water is ainly a function of P. If the tank is open to the atosphere, then P P 0 and v!2gh. In other words, for an open tank, the speed of liquid coing out through a hole a distance h below the surface is equal to that acquired by an object falling freely through a vertical distance h. This phenoenon is known as Torricelli s law. (a) ptional Section 5.8 THER APPLICATINS F BERNULLI S EQUATIN (b) F Figure 5.24 Strealine flow around an airplane wing. The pressure above the wing is less than the pressure below, and a dynaic lift upward results. The lift on an aircraft wing can be eplained, in part, by the Bernoulli effect. Airplane wings are designed so that the air speed above the wing is greater than that below the wing. As a result, the air pressure above the wing is less than the pressure below, and a net upward force on the wing, called lift, results. Another factor influencing the lift on a wing is shown in Figure The wing has a slight upward tilt that causes air olecules striking its botto to be deflected downward. This deflection eans that the wing is eerting a downward force on the air. According to Newton s third law, the air ust eert an equal and opposite force on the wing. Finally, turbulence also has an effect. If the wing is tilted too uch, the flow of air across the upper surface becoes turbulent, and the pressure difference across the wing is not as great as that predicted by Bernoulli s equation. In an etree case, this turbulence ay cause the aircraft to stall. In general, an object oving through a fluid eperiences lift as the result of any effect that causes the fluid to change its direction as it flows past the object. Soe factors that influence lift are the shape of the object, its orientation with respect to the fluid flow, any spinning otion it ight have, and the teture of its surface. For eaple, a golf ball struck with a club is given a rapid backspin, as shown in Figure 5.25a. The diples on the ball help entrain the air to follow the curvature of the ball s surface. This effect is ost pronounced on the top half of the ball, where the ball s surface is oving in the sae direction as the air flow. Figure 5.25b shows a thin layer of air wrapping part way around the ball and being deflected downward as a result. Because the ball pushes the air down, the air ust push up on the ball. Without the diples, the air is not as well entrained, Figure 5.25 (a) A golf ball is ade to spin when struck by the club. (b) The spinning ball eperiences a lifting force that allows it to travel uch farther than it would if it were not spinning. and the golf ball does not travel as far. For the sae reason, a tennis ball s fuzz helps the spinning ball grab the air rushing by and helps deflect it. A nuber of devices operate by eans of the pressure differentials that result fro differences in a fluid s speed. For eaple, a strea of air passing over one end of an open tube, the other end of which is iersed in a liquid, reduces the pressure above the tube, as illustrated in Figure This reduction in pressure causes the liquid to rise into the air strea. The liquid is then dispersed into a fine spray of droplets. You ight recognize that this so-called atoizer is used in perfue bottles and paint sprayers. The sae principle is used in the carburetor of a gasoline engine. In this case, the low-pressure region in the carburetor is produced by air drawn in by the piston through the air filter. The gasoline vaporizes in that region, ies with the air, and enters the cylinder of the engine, where cobustion occurs. Quick Quiz 5.0 People in buildings threatened by a tornado are often told to open the windows to iniize daage. Why? Figure 5.26 A strea of air passing over a tube dipped into a liquid causes the liquid to rise in the tube.

117 476 CHAPTER 5 Fluid Mechanics Questions 477 QUESTINS SUMMARY. Two drinking glasses of the sae weight but of different shape and different cross-sectional area are filled to the sae level with water. According to the epression P P 0 gh, the pressure at the botto of both glasses is the sae. In view of this, why does one glass weigh ore than the other? 2. If the top of your head has a surface area of 00 c 2, what is the weight of the air above your head? 3. When you drink a liquid through a straw, you reduce the The pressure P in a fluid is the force per unit area eerted by the fluid on a surface: P F (5.) A In the SI syste, pressure has units of newtons per square eter (N/ 2 ), and N/ 2 pascal (Pa). The pressure in a fluid at rest varies with depth h in the fluid according to the epression P P 0 gh (5.4) where P 0 is atospheric pressure ( N/ 2 ) and is the density of the fluid, assued unifor. Pascal s law states that when pressure is applied to an enclosed fluid, the pressure is transitted undiinished to every point in the fluid and to every point on the walls of the container. When an object is partially or fully suberged in a fluid, the fluid eerts on the object an upward force called the buoyant force. According to Archiedes s principle, the agnitude of the buoyant force is equal to the weight of the fluid displaced by the object. Be sure you can apply this principle to a wide variety of situations, including sinking objects, floating ones, and neutrally buoyant ones. You can understand various aspects of a fluid s dynaics by assuing that the fluid is nonviscous and incopressible and that the fluid s otion is a steady flow with no rotation. Two iportant concepts regarding ideal fluid flow through a pipe of nonunifor size are as follows:. The flow rate (volue flu) through the pipe is constant; this is equivalent to stating that the product of the cross-sectional area A and the speed v at any point is a constant. This result is epressed in the equation of continuity: A v A 2 v 2 constant (5.7) You can use this epression to calculate how the velocity of a fluid changes as the fluid is constricted or as it flows into a ore open area. 2. The su of the pressure, kinetic energy per unit volue, and gravitational potential energy per unit volue has the sae value at all points along a strealine. This result is suarized in Bernoulli s equation: P 2v 2 gy constant (5.9) pressure in your outh and let the atosphere ove the liquid. Eplain why this is so. Can you use a straw to sip a drink on the Moon? 4. A heliu-filled balloon rises until its density becoes the sae as that of the surrounding air. If a sealed subarine begins to sink, will it go all the way to the botto of the ocean or will it stop when its density becoes the sae as that of the surrounding water? 5. A fish rests on the botto of a bucket of water while the bucket is being weighed. When the fish begins to swi around, does the weight change? 6. Does a ship ride higher in the water of an inland lake or in the ocean? Why? 7. Lead has a greater density than iron, and both etals are denser than water. Is the buoyant force on a lead object greater than, less than, or equal to the buoyant force on an iron object of the sae volue? 8. The water supply for a city is often provided by reservoirs built on high ground. Water flows fro the reservoir, through pipes, and into your hoe when you turn the tap on your faucet. Why is the flow of water ore rapid out of a faucet on the first floor of a building than it is in an apartent on a higher floor? 9. Soke rises in a chiney faster when a breeze is blowing than when there is no breeze at all. Use Bernoulli s equation to eplain this phenoenon. 0. If a Ping Pong ball is above a hair dryer, the ball can be suspended in the air colun eitted by the dryer. Eplain.. When ski jupers are airborne (Fig. Q5.), why do they bend their bodies forward and keep their hands at their sides? Figure Q5. 2. Eplain why a sealed bottle partially filled with a liquid can float. 3. When is the buoyant force on a swier greater after ehaling or after inhaling? 4. A piece of unpainted wood barely floats in a container partly filled with water. If the container is sealed and then pressurized above atospheric pressure, does the wood rise, sink, or reain at the sae level? (Hint: Wood is porous.) 5. A flat plate is iersed in a liquid at rest. For what orientation of the plate is the pressure on its flat surface unifor? 6. Because atospheric pressure is about 0 5 N/ 2 and the area of a person s chest is about 0.3 2, the force of the atosphere on one s chest is around N. In view of this enorous force, why don t our bodies collapse? 7. How would you deterine the density of an irregularly shaped rock? 8. Why do airplane pilots prefer to take off into the wind? 9. If you release a ball while inside a freely falling elevator, the ball reains in front of you rather than falling to the floor because the ball, the elevator, and you all eperience the sae downward acceleration g. What happens if you repeat this eperient with a heliu-filled balloon? (This one is tricky.) 20. Two identical ships set out to sea. ne is loaded with a cargo of Styrofoa, and the other is epty. Which ship is ore suberged? 2. A sall piece of steel is tied to a block of wood. When the wood is placed in a tub of water with the steel on top, half of the block is suberged. If the block is inverted so that the steel is underwater, does the aount of the block suberged increase, decrease, or reain the sae? What happens to the water level in the tub when the block is inverted? 22. Prairie dogs (Fig. Q5.22) ventilate their burrows by building a ound over one entrance, which is open to a strea of air. A second entrance at ground level is open to alost stagnant air. How does this construction create an air flow through the burrow? Figure Q An unopened can of diet cola floats when placed in a tank of water, whereas a can of regular cola of the sae brand sinks in the tank. What do you suppose could eplain this phenoenon? 24. Figure Q5.24 shows a glass cylinder containing four liquids of different densities. Fro top to botto, the liquids are oil (orange), water (yellow), salt water (green), and ercury (silver). The cylinder also contains, fro top to botto, a Ping Pong ball, a piece of wood, an egg, and a steel ball. (a) Which of these liquids has the lowest density, and which has the greatest? (b) What can you conclude about the density of each object?

118 478 CHAPTER 5 Fluid Mechanics Probles 479 Figure Q In Figure Q5.25, an air strea oves fro right to left through a tube that is constricted at the iddle. Three Ping Pong balls are levitated in equilibriu above the vertical coluns through which the air escapes. (a) Why is the ball at the right higher than the one in the iddle? PRBLEMS Section 5. Pressure. Calculate the ass of a solid iron sphere that has a diaeter of 3.00 c. 2. Find the order of agnitude of the density of the nucleus of an ato. What does this result suggest concerning the structure of atter? (Visualize a nucleus as protons and neutrons closely packed together. Each has ass kg and radius on the order of 0 5.) 3. A 50.0-kg woan balances on one heel of a pair of highheeled shoes. If the heel is circular and has a radius of c, what pressure does she eert on the floor? 4. The four tires of an autoobile are inflated to a gauge pressure of 200 kpa. Each tire has an area of in contact with the ground. Deterine the weight of the autoobile. 5. What is the total ass of the Earth s atosphere? (The radius of the Earth is , and atospheric pressure at the Earth s surface is N/ 2.) Section 5.2 Figure Q5.25 (b) Why is the ball at the left lower than the ball at the right even though the horizontal tube has the sae diensions at these two points? 26. You are a passenger on a spacecraft. For your cofort, the interior contains air just like that at the surface of the Earth. The craft is coasting through a very epty region of space. That is, a nearly perfect vacuu eists just outside the wall. Suddenly a eteoroid pokes a hole, saller than the pal of your hand, right through the wall net to your seat. What will happen? Is there anything you can or should do about it?, 2, 3 = straightforward, interediate, challenging = full solution available in the Student s Manual and Study Guide WEB = solution posted at = Coputer useful in solving proble = Interactive Physics = paired nuerical/sybolic probles Variation of Pressure with Depth 6. (a) Calculate the absolute pressure at an ocean depth of 000. Assue the density of seawater is 024 kg/ 3 and that the air above eerts a pressure of 0.3 kpa. (b) At this depth, what force ust the frae around a circular subarine porthole having a diaeter of 30.0 c eert to counterbalance the force eerted by the water? 7. The spring of the pressure gauge shown in Figure 5.2 has a force constant of 000 N/, and the piston has a diaeter of 2.00 c. When the gauge is lowered into water, at what depth does the piston ove in by c? 8. The sall piston of a hydraulic lift has a cross-sectional area of 3.00 c 2, and its large piston has a cross-sectional area of 200 c 2 (see Fig. 5.5a). What force ust be applied to the sall piston for it to raise a load of 5.0 kn? (In service stations, this force is usually generated with the use of copressed air.) WEB 9. What ust be the contact area between a suction cup (copletely ehausted) and a ceiling if the cup is to support the weight of an 80.0-kg student? 0. (a) A very powerful vacuu cleaner has a hose 2.86 c in diaeter. With no nozzle on the hose, what is the weight of the heaviest brick that the cleaner can lift (Fig. P5.0)? (b) A very powerful octopus uses one sucker of diaeter 2.86 c on each of the two shells of a cla in an attept to pull the shells apart. Find the greatest force that the octopus can eert in salt water 32.3 in depth. (Caution: Eperiental verification can be interesting, but do not drop a brick on your foot. Do not overheat the otor of a vacuu cleaner. Do not get an octopus ad at you.). For the cellar of a new house, a hole with vertical sides descending 2.40 is dug in the ground. A concrete foundation wall is built all the way across the width of the ecavation. This foundation wall is 0.83 away fro the front of the cellar hole. During a rainstor, drainage fro the street fills up the space in front of the concrete wall but not the cellar behind the wall. The water does not soak into the clay soil. Find the force that the water causes on the foundation wall. For coparison, the weight of the water is given by kg/ /s kn 2. A swiing pool has diensions and a flat botto. When the pool is filled to a depth of 2.00 with fresh water, what is the force caused by the water on the botto? n each end? n each side? 3. A sealed spherical shell of diaeter d is rigidly attached to a cart that is oving horizontally with an acceleration a, as shown in Figure P5.3. The sphere is nearly filled with a fluid having density and also contains one sall bubble of air at atospheric pressure. Find an epression for the pressure P at the center of the sphere. (a) Figure P5.0 (b) 4. The tank shown in Figure P5.4 is filled with water to a depth of At the botto of one of the side walls is a rectangular hatch.00 high and 2.00 wide. The hatch is hinged at its top. (a) Deterine the force that the water eerts on the hatch. (b) Find the torque eerted about the hinges Figure P5.3 Figure P Review Proble. A solid copper ball with a diaeter of 3.00 at sea level is placed at the botto of the ocean (at a depth of 0.0 k). If the density of seawater is 030 kg/ 3, by how uch (approiately) does the diaeter of the ball decrease when it reaches botto? Take the bulk odulus of copper as N/ 2. a

119 480 CHAPTER 5 Fluid Mechanics Probles 48 WEB Section 5.3 Pressure Measureents 6. Noral atospheric pressure is Pa. The approach of a stor causes the height of a ercury baroeter to drop by 20.0 fro the noral height. What is the atospheric pressure? (The density of ercury is 3.59 g/c 3.) 7. Blaise Pascal duplicated Torricelli s baroeter, using a red Bordeau wine, of density 984 kg/ 3, as the working liquid (Fig. P5.7). What was the height h of the wine colun for noral atospheric pressure? Would you epect the vacuu above the colun to be as good as that for ercury? P 0 Figure P Mercury is poured into a U-tube, as shown in Figure P5.8a. The left ar of the tube has a cross-sectional area A of 0.0 c 2, and the right ar has a cross-sectional area A 2 of 5.00 c 2. ne-hundred gras of water are then poured into the right ar, as shown in Figure P5.8b. (a) Deterine the length of the water colun A (a) A 2 Mercury h Figure P5.8 h A A 2 Water (b) in the right ar of the U-tube. (b) Given that the density of ercury is 3.6 g/c 3, what distance h does the ercury rise in the left ar? 9. A U-tube of unifor cross-sectional area and open to the atosphere is partially filled with ercury. Water is then poured into both ars. If the equilibriu configuration of the tube is as shown in Figure P5.9, with h 2.00 c, deterine the value of h. h h 2 Figure P5.9 Water Mercury Section 5.4 Buoyant Forces and Archiedes s Principle 20. (a) A light balloon is filled with of heliu. At 0 C, what is the ass of the payload that the balloon can lift? (b) In Table 5., note that the density of hydrogen is nearly one-half the density of heliu. What load can the balloon lift if it is filled with hydrogen? 2. A Styrofoa slab has a thickness of 0.0 c and a density of 300 kg/ 3. When a 75.0-kg swier is resting on it, the slab floats in fresh water with its top at the sae level as the water s surface. Find the area of the slab. 22. A Styrofoa slab has thickness h and density S. What is the area of the slab if it floats with its upper surface just awash in fresh water, when a swier of ass is on top? 23. A piece of aluinu with ass.00 kg and density kg/ 3 is suspended fro a string and then copletely iersed in a container of water (Fig. P5.23). Calculate the tension in the string (a) before and (b) after the etal is iersed. 24. A 0.0-kg block of etal easuring 2.0 c 0.0 c 0.0 c is suspended fro a scale and iersed in water, as shown in Figure P5.23b. The 2.0-c diension is vertical, and the top of the block is 5.00 c fro the surface of the water. (a) What are the forces acting on the top and on the botto of the block? (Take P N/ 2.) (b) What is the reading of the spring scale? (c) Show that the buoyant force equals the difference between the forces at the top and botto of the block. WEB T Mg (a) Scale Figure P5.23 Probles 23 and A cube of wood having a side diension of 20.0 c and a density of 650 kg/ 3 floats on water. (a) What is the distance fro the horizontal top surface of the cube to the water level? (b) How uch lead weight ust be placed on top of the cube so that its top is just level with the water? 26. To an order of agnitude, how any heliu-filled toy balloons would be required to lift you? Because heliu is an irreplaceable resource, develop a theoretical answer rather than an eperiental answer. In your solution, state what physical quantities you take as data and the values you easure or estiate for the. 27. A plastic sphere floats in water with 50.0% of its volue suberged. This sae sphere floats in glycerin with 40.0% of its volue suberged. Deterine the densities of the glycerin and the sphere. 28. A frog in a heispherical pod finds that he just floats without sinking into a sea of blue-green ooze having a density of.35 g/c 3 (Fig. P5.28). If the pod has a radius of 6.00 c and a negligible ass, what is the ass of the frog? Figure P How any cubic eters of heliu are required to lift a balloon with a 400-kg payload to a height of 8 000? (Take He 0.80 kg/ 3.) Assue that the balloon B Mg T 2 (b) aintains a constant volue and that the density of air decreases with the altitude z according to the epression air 0e z/8 000, where z is in eters and 0.25 kg/ 3 is the density of air at sea level. 30. Review Proble. A long cylindrical tube of radius r is weighted on one end so that it floats upright in a fluid having a density. It is pushed downward a distance fro its equilibriu position and then released. Show that the tube will eecute siple haronic otion if the resistive effects of the water are neglected, and deterine the period of the oscillations. 3. A bathysphere used for deep-sea eploration has a radius of.50 and a ass of kg. To dive, this subarine takes on ass in the for of seawater. Deterine the aount of ass that the subarine ust take on if it is to descend at a constant speed of.20 /s, when the resistive force on it is 00 N in the upward direction. Take kg/ 3 as the density of seawater. 32. The United States possesses the eight largest warships in the world aircraft carriers of the Niitz class and it is building one ore. Suppose that one of the ships bobs up to float.0 c higher in the water when 50 fighters take off fro it at a location where g 9.78 /s 2. The planes have an average ass of kg. Find the horizontal area enclosed by the waterline of the ship. (By coparison, its flight deck has an area of ) Section 5.5 Fluid Dynaics Section 5.6 Strealines and the Equation of Continuity Section 5.7 Bernoulli s Equation 33. (a) A water hose 2.00 c in diaeter is used to fill a 20.0-L bucket. If it takes.00 in to fill the bucket, what is the speed v at which water oves through the hose? (Note: L 000 c 3.) (b) If the hose has a nozzle.00 c in diaeter, find the speed of the water at the nozzle. 34. A horizontal pipe 0.0 c in diaeter has a sooth reduction to a pipe 5.00 c in diaeter. If the pressure of the water in the larger pipe is Pa and the pressure in the saller pipe is Pa, at what rate does water flow through the pipes? 35. A large storage tank, open at the top and filled with water, develops a sall hole in its side at a point 6.0 below the water level. If the rate of flow fro the leak is /in, deterine (a) the speed at which the water leaves the hole and (b) the diaeter of the hole. 36. Through a pipe of diaeter 5.0 c, water is puped fro the Colorado River up to Grand Canyon Village, located on the ri of the canyon. The river is at an elevation of 564, and the village is at an elevation of (a) What is the iniu pressure at which the water ust be puped if it is to arrive at the village?

120 482 CHAPTER 5 Fluid Mechanics Probles 483 (b) If are puped per day, what is the speed of the water in the pipe? (c) What additional pressure is necessary to deliver this flow? (Note: You ay assue that the acceleration due to gravity and the density of air are constant over this range of elevations.) 37. Water flows through a fire hose of diaeter 6.35 c at a rate of /s. The fire hose ends in a nozzle with an inner diaeter of 2.20 c. What is the speed at which the water eits the nozzle? 38. ld Faithful Geyser in Yellowstone National Park erupts at approiately -h intervals, and the height of the water colun reaches 40.0 (Fig. P5.38). (a) Consider the rising strea as a series of separate drops. Analyze the free-fall otion of one of these drops to deterine the speed at which the water leaves the ground. (b) Treating the rising strea as an ideal fluid in strealine flow, use Bernoulli s equation to deterine the speed of the water as it leaves ground level. (c) What is the pressure (above atospheric) in the heated underground chaber if its depth is 75? You ay assue that the chaber is large copared with the geyser s vent. v air A Mercury Figure P5.4 air flow. (Assue that the air is stagnant at point A, and take air.25 kg/ 3.) 42. An airplane is cruising at an altitude of 0 k. The pressure outside the craft is at; within the passenger copartent, the pressure is.00 at and the teperature is 20 C. A sall leak occurs in one of the window seals in the passenger copartent. Model the air as an ideal fluid to find the speed of the strea of air flowing through the leak. 43. A siphon is used to drain water fro a tank, as illustrated in Figure P5.43. The siphon has a unifor diaeter. Assue steady flow without friction. (a) If the distance h.00, find the speed of outflow at the end of the siphon. (b) What is the liitation on the height of the top of the siphon above the water surface? (For the flow of liquid to be continuous, the pressure ust not drop below the vapor pressure of the liquid.) h F h A hole is punched at a height h in the side of a container of height h 0. The container is full of water, as shown in Figure P5.45. If the water is to shoot as far as possible horizontally, (a) how far fro the botto of the container should the hole be punched? (b) Neglecting frictional losses, how far (initially) fro the side of the container will the water land? ADDITINAL PRBLEMS 47. A Ping Pong ball has a diaeter of 3.80 c and an average density of g/c 3. What force would be required to hold it copletely suberged under water? 48. Figure P5.48 shows a tank of water with a valve at the botto. If this valve is opened, what is the aiu height attained by the water strea eiting the right side of the tank? Assue that h 0.0, L 2.00, and, and that the cross-sectional area at point A is very large copared with that at point B. A Figure P5.44 Figure P5.45 Probles 45 and 46. h a v 49. A heliu-filled balloon is tied to a long, kg unifor string. The balloon is spherical with a radius of When released, the balloon lifts a length h of string and then reains in equilibriu, as shown in Figure P5.49. Deterine the value of h. The envelope of the balloon has a ass of kg. h 50. Water is forced out of a fire etinguisher by air pressure, as shown in Figure P5.50. How uch gauge air pressure in the tank (above atospheric) is required for the water jet to have a speed of 30.0 /s when the water level is below the nozzle? He Figure P5.49 Figure P5.38 (ptional) Section 5.8 ther Applications of Bernoulli s Equation 39. An airplane has a ass of kg, and each wing has an area of During level flight, the pressure on the lower wing surface is Pa. Deterine the pressure on the upper wing surface. 40. A Venturi tube ay be used as a fluid flow eter (see Fig. 5.2). If the difference in pressure is P P kpa, find the fluid flow rate in cubic eters per second, given that the radius of the outlet tube is.00 c, the radius of the inlet tube is 2.00 c, and the fluid is gasoline ( kg/ 3 ). 4. A Pitot tube can be used to deterine the velocity of air flow by easuring the difference between the total pressure and the static pressure (Fig. P5.4). If the fluid in the tube is ercury, whose density is Hg kg/ 3, and if h 5.00 c, find the speed of 700 WEB ρ Figure P A hypoderic syringe contains a edicine with the density of water (Fig. P5.44). The barrel of the syringe has a cross-sectional area A , and the needle has a cross-sectional area a In the absence of a force on the plunger, the pressure everywhere is at. A force F of agnitude 2.00 N acts on the plunger, aking the edicine squirt horizontally fro the needle. Deterine the speed of the edicine as it leaves the needle s tip. 45. A large storage tank is filled to a height h 0. The tank is punctured at a height h above the botto of the tank (Fig. P5.45). Find an epression for how far fro the tank the eiting strea lands. y h v A Valve 30.0 h Figure P5.48 L θ B Figure P The true weight of an object is easured in a vacuu, where buoyant forces are absent. An object of volue V is weighed in air on a balance with the use of weights of density. If the density of air is air and the balance reads F g, show that the true weight F g is F g F g V F g airg g 52. Evangelista Torricelli was the first to realize that we live at the botto of an ocean of air. He correctly surised that the pressure of our atosphere is attributable to the weight of the air. The density of air at 0 C at the Earth s surface is.29 kg/ 3. The density decreases with increasing altitude (as the atosphere thins). n the other hand, if we assue that the density is constant v

121 484 CHAPTER 5 Fluid Mechanics Probles 485 (.29 kg/ 3 ) up to soe altitude h, and zero above that altitude, then h would represent the thickness of our atosphere. Use this odel to deterine the value of h that gives a pressure of.00 at at the surface of the Earth. Would the peak of Mt. Everest rise above the surface of such an atosphere? 53. A wooden dowel has a diaeter of.20 c. It floats in water with c of its diaeter above water level (Fig. P5.53). Deterine the density of the dowel c 0.80 c Figure P A light spring of constant k 90.0 N/ rests vertically on a table (Fig. P5.54a). A 2.00-g balloon is filled with heliu (density 0.80 kg/ 3 ) to a volue of and is then connected to the spring, causing it to stretch as shown in Figure P5.54b. Deterine the etension distance L when the balloon is in equilibriu. Figure P5.55 Probles 55 and 56. da about an ais through is 6 gwh 3. Show that the effective line of action of the total force eerted by the water is at a distance 3 H above. 58. In about 657 tto von Guericke, inventor of the air pup, evacuated a sphere ade of two brass heispheres. Two teas of eight horses each could pull the heispheres apart only on soe trials, and then with greatest difficulty, with the resulting sound likened to a cannon firing (Fig. P5.58). (a) Show that the force F required to pull the evacuated heispheres apart is R 2 (P 0 P), where R is the radius of the heispheres and P is the pressure inside the heispheres, which is uch less than P 0. (b) Deterine the force if P 0.00P 0 and R In 983 the United States began coining the cent piece out of copper-clad zinc rather than pure copper. The ass of the old copper cent is g, whereas that of the new cent is 2.57 g. Calculate the percent of zinc (by volue) in the new cent. The density of copper is g/c 3, and that of zinc is 7.33 g/c 3. The new and old coins have the sae volue. 60. A thin spherical shell with a ass of 4.00 kg and a diaeter of is filled with heliu (density 0.80 kg/ 3 ). It is then released fro rest on the botto of a pool of water that is 4.00 deep. (a) Neglecting frictional effects, show that the shell rises with constant acceleration and deterine the value of that acceleration. (b) How long does it take for the top of the shell to reach the water s surface? 6. An incopressible, nonviscous fluid initially rests in the vertical portion of the pipe shown in Figure P5.6a, where L When the valve is opened, the fluid flows into the horizontal section of the pipe. What is the speed of the fluid when all of it is in the horizontal section, as in Figure P5.6b? Assue that the cross-sectional area of the entire pipe is constant. Figure P5.63 WEB k (a) Figure P A.00-kg beaker containing 2.00 kg of oil (density 96.0 kg/ 3 ) rests on a scale. A 2.00-kg block of iron is suspended fro a spring scale and copletely suberged in the oil, as shown in Figure P5.55. Deterine the equilibriu readings of both scales. 56. A beaker of ass b containing oil of ass 0 (density 0 ) rests on a scale. A block of iron of ass Fe is suspended fro a spring scale and copletely suberged in the oil, as shown in Figure P5.55. Deterine the equilibriu readings of both scales. 57. Review Proble. With reference to Figure 5.7, show that the total torque eerted by the water behind the L k (b) Figure P5.58 F P The colored engraving, dated 672, illustrates tto von Guericke s deonstration of the force due to air pressure as perfored before Eperor Ferdinand III in 657. ( R P 0 F L Valve closed (a) Figure P5.6 Valve opened 62. Review Proble. A unifor disk with a ass of 0.0 kg and a radius of spins at 300 rev/in on a low-friction ale. It ust be brought to a stop in.00 in by a brake pad that akes contact with the disk at an average distance of fro the ais. The coefficient of friction between the pad and the disk is A piston in a cylinder with a diaeter of 5.00 c presses the brake pad against the disk. Find the pressure that the brake fluid in the cylinder ust have. 63. Figure P5.63 shows Superan attepting to drink water through a very long straw. With his great strength, L (b) v he achieves aiu possible suction. The walls of the tubular straw do not collapse. (a) Find the aiu height through which he can lift the water. (b) Still thirsty, the Man of Steel repeats his attept on the Moon, which has no atosphere. Find the difference between the water levels inside and outside the straw. 64. Show that the variation of atospheric pressure with altitude is given by P P 0 e h, where 0g /P 0, P 0 is atospheric pressure at soe reference level y 0, and 0 is the atospheric density at this level. Assue that the decrease in atospheric pressure with increasing altitude is given by Equation 5.4, so that dp/dy g, and assue that the density of air is proportional to the pressure. 65. A cube of ice whose edge easures 20.0 is floating in a glass of ice-cold water with one of its faces parallel to the water s surface. (a) How far below the water surface is the botto face of the block? (b) Ice-cold ethyl alcohol is gently poured onto the water s surface to for a layer 5.00 thick above the water. The alcohol does not i with the water. When the ice cube again attains hydrostatic equilibriu, what is the distance fro the top of the water to the botto face of the block? (c) Additional cold ethyl alcohol is poured onto the water s surface until the top surface of the alcohol coincides with the top surface of the ice cube (in

122 486 CHAPTER 5 Fluid Mechanics Answers to Quick Quizzes 487 hydrostatic equilibriu). How thick is the required layer of ethyl alcohol? 66. Review Proble. A light balloon filled with heliu with a density of 0.80 kg/ 3 is tied to a light string of length L The string is tied to the ground, foring an inverted siple pendulu, as shown in Figure P5.66a. If the balloon is displaced slightly fro its equilibriu position as shown in Figure P5.66b, (a) show that the ensuing otion is siple haronic and (b) deterine the period of the otion. Take the density of air to be.29 kg/ 3 and ignore any energy loss due to air friction. Air g (a) L He Air g Figure P The water supply of a building is fed through a ain 6.00-c-diaeter pipe. A 2.00-c-diaeter faucet tap located 2.00 above the ain pipe is observed to fill a 25.0-L container in 30.0 s. (a) What is the speed at which the water leaves the faucet? (b) What is the gauge pressure in the 6-c ain pipe? (Assue that the faucet is the only leak in the building.) 68. The spirit-in-glass theroeter, invented in Florence, Italy, around 654, consists of a tube of liquid (the spirit) containing a nuber of suberged glass spheres with slightly different asses (Fig. P5.68). At sufficiently low teperatures, all the spheres float, but as the teperature rises, the spheres sink one after the other. The device is a crude but interesting tool for easuring teperature. Suppose that the tube is filled with ethyl alcohol, whose density is g/c 3 at 20.0 C and decreases to g/c 3 at 30.0 C. (a) If one of the spheres has a radius of.000 c and is in equilibriu halfway up the tube at 20.0 C, deterine its ass. (b) When the teperature increases to 30.0 C, what ass ust a second sphere of the sae radius have to be in equilibriu at the halfway point? (c) At 30.0 C the first sphere has fallen to the botto of the tube. What upward force does the botto of the tube eert on this sphere? 69. A U-tube open at both ends is partially filled with water (Fig. P5.69a). il having a density of 750 kg/ 3 is then poured into the right ar and fors a colun L 5.00 c in height (Fig. P5.69b). (a) Deterine θ (b) L He Water P 0 Figure P5.68 the difference h in the heights of the two liquid surfaces. (b) The right ar is shielded fro any air otion while air is blown across the top of the left ar until the surfaces of the two liquids are at the sae height (Fig. P5.69c). Deterine the speed of the air being blown across the left ar. (Take the density of air as.29 kg/ 3.) h il (a) (b) (c) L Figure P5.69 v Shield L ANSWERS T QUICK QUIZZES 5. You would be better off with the basketball player. Although weight is distributed over the larger surface area, equal to about half of the total surface area of the sneaker sole, the pressure (F/A) that he applies is relatively sall. The woan s lesser weight is distributed over the very sall cross-sectional area of the spiked heel. Soe useus ake woen in high-heeled shoes wear slippers or special heel attachents so that they do not daage the wood floors. 5.2 If the professor were to try to support his entire weight on a single nail, the pressure eerted on his skin would be his entire weight divided by the very sall surface area of the nail point. This etreely great pressure would cause the nail to puncture his skin. However, if the professor distributes his weight over several hundred nails, as shown in the photograph, the pressure eerted on his skin is considerably reduced because the surface area that supports his weight is now the total surface area of all the nail points. (Lying on the bed of nails is uch ore cofortable than sitting on the bed, and standing on the bed without shoes is definitely not recoended. Do not lie on a bed of nails unless you have been shown how to do so safely.) 5.3 Because the horizontal force eerted by the outside fluid on an eleent of the cylinder is equal and opposite the horizontal force eerted by the fluid on another eleent diaetrically opposite the first, the net force on the cylinder in the horizontal direction is zero. 5.4 If you think of the grain stored in the silo as a fluid, then the pressure it eerts on the walls increases with increasing depth. The spacing between bands is saller at the lower portions so that the greater outward forces acting on the walls can be overcoe. The silo on the right shows another way of accoplishing the sae thing: double banding at the botto. 5.5 Because water is so uch less dense than ercury, the colun for a water baroeter would have to be h P 0 /g 0.3 high, and such a colun is inconveniently tall. 5.6 The entire hull of a ship is full of air, and the density of air is about one-thousandth the density of water. Hence, the total weight of the ship equals the weight of the volue of water that is displaced by the portion of the ship that is below sea level. 5.7 Reains the sae. In effect, the ice creates a hole in the water, and the weight of the water displaced fro the hole is the sae as all the weight of the cube. When the cube changes fro ice to water, the water just fills the hole. 5.8 Goes down because the anchor displaces ore water while in the boat than it does in the pond. While it is in the boat, the anchor can be thought of as a floating object that displaces a volue of water weighing as uch as it does. When the anchor is thrown overboard, it sinks and displaces a volue of water equal to its own volue. Because the density of the anchor is greater than that of water, the volue of water that weighs the sae as the anchor is greater than the volue of the anchor. 5.9 As the water falls, its speed increases. Because the flow rate Av ust reain constant at all cross sections (see Eq. 5.7), the strea ust becoe narrower as the speed increases. 5.0 The rapidly oving air characteristic of a tornado is at a pressure below atospheric pressure. The stationary air inside the building reains at atospheric pressure. The pressure difference results in an outward force on the roof and walls, and this force can be great enough to lift the roof off the building. pening the windows helps to equalize the inside and outside pressures.

123 P U Z Z L E R A siple seisograph can be constructed with a spring-suspended pen that draws a line on a slowly unrolling strip of paper. The paper is ounted on a structure attached to the ground. During an earthquake, the pen reains nearly stationary while the paper shakes beneath it. How can a few jagged lines on a piece of paper allow scientists at a seisograph station to deterine the distance to the origin of an earthquake? (Ken M. Johns/Photo Researchers, Inc.) c h a p t e r Wave Motion Most of us eperienced waves as children when we dropped a pebble into a pond. At the point where the pebble hits the water s surface, waves are created. These waves ove outward fro the creation point in epanding circles until they reach the shore. If you were to eaine carefully the otion of a leaf floating on the disturbed water, you would see that the leaf oves up, down, and sideways about its original position but does not undergo any net displaceent away fro or toward the point where the pebble hit the water. The water olecules just beneath the leaf, as well as all the other water olecules on the pond s surface, behave in the sae way. That is, the water wave oves fro the point of origin to the shore, but the water is not carried with it. An ecerpt fro a book by Einstein and Infeld gives the following rearks concerning wave phenoena: A bit of gossip starting in Washington reaches New York [by word of outh] very quickly, even though not a single individual who takes part in spreading it travels between these two cities. There are two quite different otions involved, that of the ruor, Washington to New York, and that of the persons who spread the ruor. The wind, passing over a field of grain, sets up a wave which spreads out across the whole field. Here again we ust distinguish between the otion of the wave and the otion of the separate plants, which undergo only sall oscillations... The particles constituting the ediu perfor only sall vibrations, but the whole otion is that of a progressive wave. The essentially new thing here is that for the first tie we consider the otion of soething which is not atter, but energy propagated through atter. The world is full of waves, the two ain types being echanical waves and electroagnetic waves. We have already entioned eaples of echanical waves: sound waves, water waves, and grain waves. In each case, soe physical ediu is being disturbed in our three particular eaples, air olecules, water olecules, and stalks of grain. Electroagnetic waves do not require a ediu to propagate; soe eaples of electroagnetic waves are visible light, radio waves, television signals, and -rays. Here, in Part 2 of this book, we study only echanical waves. The wave concept is abstract. When we observe what we call a water wave, what we see is a rearrangeent of the water s surface. Without the water, there would be no wave. A wave traveling on a string would not eist without the string. Sound waves could not travel through air if there were no air olecules. With echanical waves, what we interpret as a wave corresponds to the propagation of a disturbance through a ediu. 6. Wave Motion 49 Chapter utline Basic Variables of Wave Motion 6.2 Direction of Particle Displaceent 6.3 ne-diensional Traveling Waves 6.4 Superposition and Interference 6.5 The Speed of Waves on Strings 6.6 Reflection and Transission 6.7 Sinusoidal Waves 6.8 Rate of Energy Transfer by Sinusoidal Waves on Strings 6.9 (ptional) The Linear Wave Equation Interference patterns produced by outwardspreading waves fro any drops of liquid falling into a body of water. A. Einstein and L. Infeld, The Evolution of Physics, New York, Sion & Schuster, 96. Ecerpt fro What Is a Wave?